Motion in a plane

Equation of trajectory:

Consider a point \(P\left( x,y \right)\) at time \(t\).

Horizontal motion:

Since acceleration due to gravity acts along the vertical and hence, has got no component along the horizontal i.e.,\({{a}_{x}}=0\) . So,horizontal motion is a non-accelerated motion with uniform velocity.

\(\Rightarrow x={{u}_{x}}t=ut\)

Vertical motion:

Also \({{a}_{y}}=g\,\, and \,\,{{u}_{y}}=0\)

Since, \(y={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}\)

\(\Rightarrow \,y=0+\frac{1}{2}g{{t}^{2}}\)

From (1), we get \(t=\frac{x}{u}\)

Substituting in (2), we get \(y=\left( \frac{g}{2{{u}^{2}}} \right){{x}^{2}}\)

Which is the equation of a parabola.

Velocity at any instant (t):

Horizontal motion:

Since horizontal motion is non-accelerated,\(\Rightarrow \,\,{{v}_{x}}={{u}_{x}}=u\)

\(\Rightarrow \,\,{{v}_{y}}=0+gt\)

\(\Rightarrow \,\,{{v}_{y}}=gt\)

Since, we have

 \(​​​​\bar{v}={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j}\)      

\(\Rightarrow \,\,\bar{v}=u\hat{i}+\left( gt \right)\hat{j}\)

\(\Rightarrow \,\,\left| {\bar{v}} \right|=v=\sqrt{{{u}^{2}}+{{g}^{2}}{{t}^{2}}}\)

If \(\beta \)is the angle made by \(\bar{v}\) with the horizontal, then \(\tan \beta =\frac{{{v}_{y}}}{{{v}_{x}}}=\frac{gt}{u}\)

If  \(h\) is the distance of the ground from the point of launch, T is the time taken to strike the ground and R is the range of the projectile when it hits the ground, then

\(h=\frac{1}{2}8{{T}^{2}}\)

\(\Rightarrow T=\sqrt{\frac{2h}{g}}\) and \(R=uT\)

\(\Rightarrow \,R=u\sqrt{\frac{2h}{g}}\)

Deviation suffered by a horizontal projectile in time “t”:

If gravity were absent then the body launched with initial velocity \(u\) would continue  to move horizontal  forever. However in the presence of gravity it suffers a deviation 

\(\delta =y=\frac{1}{2}g{{t}^{2}}=\left( \frac{g}{2{{u}^{2}}} \right){{x}^{2}}\)    

1. A body when projected vertically up with initial velocity v₀ covers a total distance D during its time of flight. If there were no gravity, the distance covered by it during the same time ( velocity equal to v₀) is equal to

Solution: 

The displacement of the body during the time t as it comes back to the point of projection

S = 0

v₀t -\(\frac{1}{2}\) gt² = 0  \(\rightarrow\) \(t=\frac{2v_0}{g}\)

During the same time t, the body moves in absence of gravity through a distance D' = v.t, because in absence of gravity g = 0

D' =\(t=v_0\times\frac{2v_0}{g}=\frac{2v_0^2}{g}\)                       …(1)

In presence of gravity the total distance covered is

D = 2H = \(2\frac{v_0^2}{2g}=\frac{v_0^2}{g}\)                             …(2)

D' = 2D

1. A man is running with constant speed of \(v=10\sqrt{5}\,\,m/s\)on a horizontal track of radius \(R=20\,m\)as shown in figure. At position A man launches a stone in space (without changing his own speed) so that he can catch stone at B (diametrically opposite to A). The speed of launch will be approximately 4K m/sec. Then find the value of K. (take \({{\pi }^{\,2}}=10\,\,\text{and}\,\,\,g=10\,m/{{s}^{2}}\)) (x-y plane is horizontal)

Solution: 

\(t=\pi\,\,R/v=2u\,\,sin\theta/g\,\,\,\,.......(i)\\ and\,\,(ucos\theta)t=2R\,\,\,\, \,\,\,\,\,\,\,\,\,.......(ii)\)

from (i) and (ii)

u=20 m/s

2. Distance between a frog and an insect on a horizontal plane is 9 m. Frog can jump with a maximum speed of \(\sqrt{10}m/s\). Minimum number of jumps required by the frog to catch the insect is (Take g = 10 \(m/{{s}^{2}}\))

Solution: 

Maximum distance covered in one jump \(=\frac{{{u}^{2}}}{g}=1m\)

So minimum no. of jumps = 9

\(OB=10\cos \theta \,=10\times \frac{3}{\sqrt{10}}=3\sqrt{10}m\)

\(t=\frac{3\sqrt{10}}{\sqrt{10}}=3\sec \)

1. A particle  is projected horizontally with a speed \(v=5m/s\) from the top of a plane inclined at an angle\(\theta ={{37}^{0}}\) to the horizontal as shown in the figure. 

1. How far from the point of projection will the particle strike the plane?

\(R={{u}_{x}}t+\frac{1}{2}{{a}_{x}}{{t}^{2}}=v\cos \theta t+\frac{1}{2}g\sin \theta {{t}^{2}}\)

\(5\cos {{37}^{0}}\times \frac{3}{4}+\frac{1}{2}\times 10\times \sin {{37}^{0}}\times {{\left( \frac{3}{4} \right)}^{2}}=\frac{75}{16}m\)

2. Find the time taken by the particle to hit the plane

\({{a}_{x}}-g\sin \theta ,{{a}_{y}}=-g\cos \theta\)

\({{S}_{y}}=0\Rightarrow {{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}=0\Rightarrow v\sin \theta t-\frac{1}{2}g\cos \theta {{t}^{2}}=0\)

\(\Rightarrow t=\frac{2v\sin \theta }{g\cos \theta }=\frac{3}{4}s\)

3. What is the velocity of the particle just before it hits the plane?

\({{v}_{x}}={{u}_{x}}+{{a}_{x}}t=5\cos {{37}^{0}}+g\sin {{37}^{0}}\times \frac{3}{4}\)

\({{v}_{y}}={{u}_{y}}+{{a}_{y}}t=5\sin {{37}^{0}}-g\cos {{37}^{0}}\times \frac{3}{4}=-3m/s\)

\(v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{\left( \frac{17}{2} \right)}^{2}}+{{3}^{2}}}=\frac{5}{2}\sqrt{13}m/s\)  

 1. Points A and C are on the horizontal ground and A and B are in same vertical line at a separation of 1500m. Simultaneously bullets are fired from A, B and C and they collide at O. The bullet at B is fired at an angle of 30° with horizontal towards the ground at velocity 100 m/s. The bullet at C is projected vertically upward at velocity of 100 m/s. The bullet projected from A reaches its maximum height at O.

1. Find the time in which bullets will collide (seconds)

For bullet B, vertical distance travelled \(=u\sin {{30}^{o}}t+\frac{1}{2}g{{t}^{2}}\)

For bullet C, vertical distance travelled

\(=u't-\frac{1}{2}g{{t}^{2}}\)

For equations (1) and (2),

\(u\sin {{30}^{o}}t+u't=1500\)

\(50t+100t=1500\)

\(t=10\sec\)

2.Find the elevation angle \(\angle \phi =\angle OAC\)

CO = greatest height for bullet A

\(=u't-\frac{1}{2}g{{t}^{2}}=100\left( 10 \right)-\frac{1}{2}{{\left( 10 \right)}^{3}}=500\)

\(AC=\frac{1}{2}\) (Range for bullet at A)

\(=u\cos {{30}^{o}}t=500\sqrt{3}\)

\(\tan \phi =\frac{CO}{AC}=\frac{1}{\sqrt{3}}\)

\(\phi=30^0\)

3. Find the velocity of bullet at A

If q is the angle of projection at A

\(\tan \theta =2\tan \phi =\frac{2}{\sqrt{3}}\)

\(\cos \theta =\sqrt{\frac{3}{7}}\)

For bullet at A,\({{u}_{A}}\cos \theta t=500\sqrt{3}\)

\({{u}_{A}}=\frac{500\sqrt{3}\left( \sqrt{7} \right)}{\sqrt{3}\times 10}=50\sqrt{7}\,m/s\)

Then the body launched with initial velocity would continue to move forever along the line OA. However gravity is present due to which it suffers a deviation \(\delta\) form its actual track (in the absence of gravity).

In triangle OAN

\(\tan \theta =\frac{AN}{ON}=\frac{y+\delta }{x}\)

\(\Rightarrow \delta =x\tan \theta -y\)

Now since we know that \(y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\)

\(\delta =\left( x\tan \theta \right)-\left( x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta } \right)\)

\(\Rightarrow \delta =\frac{1}{2}g{{\left( \frac{x}{u\cos \theta } \right)}^{2}}\)

Since ,\(x=\left( u\cos \theta \right)t\)

\(\Rightarrow \delta =\frac{1}{2}g{{t}^{2}}\)

Hence  deviation suffered by an oblique  projectile in time t ( or in terms of x ,u and launch angle \(\theta\) ) is

\(\delta =\frac{1}{2}g{{t}^{2}}=\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\)

Particle projected at an angle \(\theta\) above the horizontal

The situation is shown in figure. In this case we have

\(y=-h,{{u}_{x}}=u\cos \theta ,{{u}_{y}}=u\sin \theta\) and \({{a}_{y}}=-g\)

For horizontal motion \(x=\left( u\cos \theta \right)t\)

For vertical motion – h = \(\left( u\sin \theta \right)t-\frac{1}{2}g{{t}^{2}}\)

\(\Rightarrow g{{t}^{2}}-\left( 2u\sin \theta \right)t-2h=0\)

\(\Rightarrow t=\frac{u\sin \theta }{g}\pm \sqrt{\frac{{{u}^{2}}{{\sin }^{2}}\theta }{{{g}^{2}}}+\frac{2h}{g}}\)

1. A small body is dropped from a rising balloon. A person ‘A’ stands on ground, while another person ‘B’ is on the balloon. Choose the correct statement: Immediately. After the body is released.

When the body is dropped from the balloon, it also acquires the upward velocity of balloon. So w.r.t a person on the ground, the ball appears to be going up. But a person on the balloon is also going up.So w.r.t him the velocity of the body will be zero and he will then see the body to be coming down.

2. A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with a speed of \(20m{{s}^{-1}}\). Two seconds later, an identical stone is thrown vertically downward with the same speed \(20m{{s}^{-1}}\). [Neglect the air resistance ]Then

Solution: 

\(\begin{align} & {{v}_{1}}=20-gt \\ & {{v}_{2}}=20-g(t-2) \\ & {{v}_{1}}-{{v}_{2}}=[20-gt]-[(20-g)(t-2)] \\ & or\,\,\,\,\,\,\,\,\,{{v}_{1}}-{{v}_{2}}=-gt+g(t-2) \\ & or\,\,\,\,\,\,\,\,\,{{v}_{1}}-{{v}_{2}}=-2g \\ \end{align}\)

Clearly, it is time – independent.

1. A body is thrown vertically upwards from the top of a tower. It reaches the ground in time \({{t}_{1}}\). If it is thrown vertically downwards from the same point with the same speed, it reaches the ground in time \({{t}_{2}}\). If it is allowed to fall from the same point then the time it takes to reach the ground is

Solution: 

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then \(u\cos {{60}^{0}}=v\cos {{45}^{0}}\)

\(\Rightarrow 150\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\) or \(v=\frac{150}{\sqrt{2}}m/s\)

Initially: \({{u}_{y}}=u\sin {{60}^{0}}=\frac{150\sqrt{3}}{2}m/s\)

Finally: \({{v}_{y}}=v\sin {{45}^{0}}=\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}m/s\)

But \({{v}_{y}}={{u}_{ y}}+{{a}_{y}}t\Rightarrow \frac{150}{2}=\frac{150\sqrt{3}}{2}-10t\)

\(10t=\frac{150}{2}(\sqrt{3}-1)\Rightarrow t=7.5(\sqrt{3}-1)s\)

2. Velocity of a particle of mass 2 kg changes from \(\overrightarrow{{{v}_{1}}}=-2\widehat{i}-2\widehat{j}\ m/s\)to \(\overrightarrow{{{v}_{2}}}=\left( \widehat{i}-\widehat{j} \right)\ m/s\) after colliding with a plane surface.

Solution: 

\(2\sqrt{2}\cos \left( 45+\theta \right)=\sqrt{2}\cos \left( 45-\theta \right)\)

\(2\left[ \frac{1}{\sqrt{2}}c\theta -\frac{1}{\sqrt{2}}s\theta \right]=\frac{1}{\sqrt{2}}c\theta +\frac{1}{\sqrt{2}}s\theta\)

\(c\theta =3s\theta \Rightarrow \tan \theta =\frac{1}{3}\)

1. A particle takes 9s and 4s respectively to reach foot of the building when projected vertically upwards and downwards respectively with same speed from top of the building. How much time (in s) particle will take to reach foot of the building when it is dropped from top of the building. 

Solution: 

\(h=-u{{t}_{1}}+{\scriptstyle{}^{1}/{}_{2}}g\,t_{1}^{2}\)

\(h=-u{{t}_{2}}+{\scriptstyle{}^{1}/{}_{2}}g\,t_{2}^{2}\)

\(\Rightarrow h{{t}_{2}}=-4{{t}_{1}}{{t}_{2}}+{\scriptstyle{}^{1}/{}_{2}}g\,t_{1}^{2}{{t}_{2}}\)

\(h{{t}_{1}}=u{{t}_{1}}{{t}_{2}}+{\scriptstyle{}^{1}/{}_{2}}g\,t_{2}^{2}{{t}_{1}}\)

\(h={\scriptstyle{}^{1}/{}_{2}}g\,t_{1}^{{}}{{t}_{2}}\)

\({\scriptstyle{}^{1}/{}_{2}}g\,{{t}^{2}}={\scriptstyle{}^{1}/{}_{2}}g{{t}_{1}}{{t}_{2}}\)

\(\therefore t=\sqrt{{{t}_{1}}{{t}_{2}}}=\sqrt{9\times 4}=6\)

2.  A particle moves along a straight line and its velocity depends on time at \(V=4\,t-{{t}^{2}}\). Then for first 5 S.

Solution: 

\(\begin{align} & \overset{\scriptscriptstyle\rightharpoonup}{V}=\frac{\int\limits_{0}^{5}{Vdt}}{\int\limits_{0}^{4}{Vdt}}=\frac{5}{3}m/s \\ & V=\frac{\left| \int\limits_{0}^{4}{Vdt} \right|+\left| \int\limits_{4}^{5}{Vdt} \right|}{\int\limits_{0}^{5}{Vdt}}=\frac{5}{3}m/s \\ & a=\frac{dV}{dT}=4-2t \\ & \\ \end{align}\)

1. Two  guns situated at the top of a hill of height  10m fire one shot each with the same  speed of \(5\sqrt{3}m{{s}^{-1}}\) at some  interval of time. First gun fires horizontally  and other (second)fires upwards  at an angle of \({{60}^{0}}\) with the  horizontal. The shots collide in mid air at the point P. Taking the origin of the coordinate system at the foot of the hill right below the muzzle trajectories  in x-y plane and  \(g=10m{{s}^{-2}}\) then

Solution: 

Let the second shell reach the point \(P\left( x,y \right)\) at time \({{t}_{2}}\) and the first shell be fired at a time

\(\left( {{t}_{2}}-\Delta t \right)=t\)

For shell 1

\(x=\left( 5\sqrt{3} \right)t\)

\(-\left( 10-y \right)=-\frac{1}{2}gt_{1}^{2}\)

For Shell 2

\(x=\left( 5\sqrt{3}\cos 60 \right){{t}_{2}}\)

\(-\left( 10-y \right)=\left( 5\sqrt{3}\sin 60 \right){{t}_{2}}-\frac{1}{2}gt_{2}^{2}\)

Equating (1) and (3), we get \({{t}_{2}}=2{{t}_{1}}\)

Equating (2) and (4) we get

\(\frac{1}{2}gt_{1}^{2}=\left( 5\sqrt{3}\sin 60 \right){{t}_{2}}-\frac{1}{2}gt_{2}^{2}\)

\(=\frac{1}{2}gt_{1}^{2}=\frac{15}{2}\left( 2{{t}_{1}} \right)-\frac{1}{2}g{{\left( 2{{t}_{1}} \right)}^{2}}\)

\(=\frac{1}{2}gt_{1}^{2}=15{{t}_{1}}-2g_{1}^{2}\)

\(\Rightarrow -15t_{1}^{2}=15{{t}_{1}}-20_{1}^{2}\)

\(\Rightarrow 15t_{1}^{2}=15{{t}_{1}}\)

\(\Rightarrow 15t_{1}^{2}-15{{t}_{1}}=0\)

\(\Rightarrow 15{{t}_{1}}\left( {{t}_{1}}-1 \right)=0\)

\(\Rightarrow {{t}_{1}}=1s\)

\(\Rightarrow {{t}_{2}}=2s i.e. \Delta t=1s\)

\(\Rightarrow x=5\sqrt{3}\text{ m }\,\, and \,\,\,y=5\text{ m}\)

Particle  projected at an angle \(\theta\) below  the horizontal

The  situation is shown  is shown in figure . In this cases  we have

\(y=+h,{{u}_{x}}=u\cos \theta ,{{u}_{y}}=u\sin \theta\)and \({{a}_{y}}=+g\)

For horizontal  motion  x =\(\left( u\cos \theta \right)t\)

For vertical motion + h= \(\left( u\sin \theta \right)t+\frac{1}{2}g{{t}^{2}}\)

\(\Rightarrow g{{t}^{2}}+\left( 2u\sin \theta \right)t-2h=0\)

\(\Rightarrow t=\frac{-u\sin \theta }{g}\pm \sqrt{\frac{{{u}^{2}}{{\sin }^{2}}\theta }{{{g}^{2}}}+\frac{2h}{g}}\)

Here negative  root should be excluded otherwise t would  be the negative

RANGE , MAXIMUM HEIGHT AND TIME OF FLIGHT FOR COMPLENTARY ANGLES

For complimentary angles say \(\phi \) and \(\left( 90-\phi \right)\), let the corresponding ranges

maximum heights and times of flight be \({{R}_{\phi }},{{R}_{90-\phi }},{{H}_{\phi }},{{H}_{90-\phi }}\)

and \({{R}_{90-\phi }}=\frac{2{{u}^{2}}\cos \phi \sin \phi }{g}\)

\(\Rightarrow {{R}_{\phi }}={{R}_{90-\phi }}=\frac{2{{u}^{2}}\sin \phi \cos \phi }{g}\)                    ………(1)

So we must remember  that  for  complimentary  angles range is the same  i.e

\({{R}_{{{1}^{0}}}}={{R}_{{{89}^{0}},}}{{R}_{{{15}^{0}}}},{{R}_{{{75}^{0}}}}\) and  so on Further more

\({{H}_{\phi }}=\frac{{{u}^{2}}{{\sin }^{2}}\phi }{2g}\)                                           …………(2)

\({{H}_{90-\phi }}=\frac{{{u}^{2}}{{\sin }^{2}}\left( 90-\phi \right)}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}\phi }{2g}\)                ………….(3)

From (2) and (3) we observe that

\({{H}_{\phi }}+{{H}_{90-\phi }}=\frac{{{u}^{2}}}{2g}\)                                 …………..(4)

Also from equations (1) (2) and (3) we observe that

\({{R}_{\phi }}={{R}_{90-\phi }}=4\sqrt{{{H}_{\phi }}{{H}_{90-\phi }}}\)                ..…………(5)

\(\Rightarrow {{R}_{{{1}^{0}}}}={{R}_{{{89}^{0}}}}=4\sqrt{{{H}_{{{1}^{0}}}}{{H}_{{{89}^{0}}}}}\)

Finally

\({{T}_{\phi }}=\frac{2u\sin \phi }{g}\,\,and\,\,{{T}_{90-\phi }}=\frac{2u\cos \phi }{g}\)     

\(\Rightarrow {{T}_{\phi }}{{T}_{90-\phi }}=\frac{4{{u}^{2}}\sin \phi \cos \phi }{{{g}^{2}}}=\frac{2}{g}\left( \frac{2{{u}^{2}}\sin \phi \cos \phi }{g} \right)\)                              

\(\Rightarrow {{T}_{\phi }}{{T}_{90-\phi }}=\frac{2{{R}_{\phi }}}{g}=\frac{2{{R}_{90-\phi }}}{g}\)     ……….(6)

i.e \({{T}_{{{1}^{0}}}}{{T}_{{{89}^{0}}}}=\frac{2{{R}_{{{1}^{0}}}}}{g}=\frac{2{{R}_{{{89}^{0}}}}}{g}\)

1. Ball A was dropped from the top of a tall building. At the same height ball B was thrown straight downward. Neglecting the effects of air friction, compare their accelerations while they were falling

Both are moving under gravity

1.Consider the situation shown in the figure. A particle has to be projected from point P in horizontal direction. It is required for the particle to strike the plane AB (see figure)

Solution: 

1. The minimum value of horizontal velocity of particle at point P so that particle may strike the plane AB is

For  minimum  horizontal  velocity  i.e.. for the  particle  just to  land  at A  we have

\(x=3a\) and \(y=4a\)

\(\Rightarrow 4a=g\frac{{{\left( 3a \right)}^{2}}}{2{{u}^{2}}_{\min }}\)

\(\Rightarrow {{u}^{2}}_{\min }=\frac{9}{8}ag\)

\(\Rightarrow {{u}_{\min }}=\sqrt{\frac{9}{8}ag}\)

2. The time taken by particle in going  from  point P to plane AB is

Since \(y=\frac{1}{2}{{a}_{y}}{{t}^{2}}\)

\(\Rightarrow 4a=\frac{1}{2}g{{t}^{2}}\)

\(\Rightarrow t=\sqrt{\frac{8a}{g}}\)

3. The maximum  value of horizontal  velocity  of particle at point  P so that it may strike plane  AB is

For maximum  horizontal  velocity i.e..  when the  particle  lands at B we have

\(x=3a+2a=5a\) and \(y=4a\)

So \(4a=\frac{g{{\left( 5a \right)}^{2}}}{2{{u}^{2}}_{\max }}\)

\(\Rightarrow {{u}_{\max }}=\sqrt{\frac{25}{8}ag}\)

1.A very  borad elevator is going  up vertically  with a constant  acceleration of \(2m{{s}^{-2}}\). At the instant when its velocity is \(4m{{s}^{-1}}\) a ball is projected from the floor of the lift with a speed of \(4m{{s}^{-1}}\) relative  to the floor at an elevation of \({{30}^{0}}\). The time taken by the ball to return the floor is \(\left( g=10m{{s}^{-2}} \right)\)

Solution: 

Let us first calculate the components of the velocity of ball relative to the lift

\({{u}_{x}}=4\cos \left( {{30}^{0}} \right)=2\sqrt{3}m{{s}^{-1}}\) and

\({{u}_{y}}=4\sin \left( {{30}^{0}} \right)=2m{{s}^{-1}}\)

The acceleration of the ball with respect to the lift is \(\left( 10+2 \right)=12m{{s}^{-2}}\) in the negative y-direction ( i.e. vertically  downwards )

So the time of flight T is

\(T=\frac{2{{u}_{y}}}{{{a}_{y}}}=\frac{2\left( 2 \right)}{12}=\frac{1}{3}s\)

A body is allowed to fall from a height of 100m. If the time taken for the first 50m is \({{t}_{1}}\) and for the remaining 50m is \({{t}_{2}}\).

1.Which is correct?

\(s=\frac{1}{2}gt_{1}^{2} \,\,or \,\,t_{1}^{2}=\frac{50\times 2}{g}=\frac{100}{g}\) or

\({{t}_{1}}=\frac{10}{\sqrt{8}}\, \,and\,\, \,100=\frac{1}{2}g{{t}^{2}}\,\,\, or \,\,\,t=\frac{10\sqrt{2}}{\sqrt{g}}\)

\({{t}_{2}}=t-{{t}_{1}}=\frac{10}{\sqrt{g}}(\sqrt{2}-1)=0.4{{t}_{1}}\)

\({{t}_{1}}>{{t}_{2}}\)

2. The ratio of \({{t}_{1}}\) and \({{t}_{2}}\) is nearly

\({{t}_{1}}:{{t}_{2}}=\frac{10}{\sqrt{g}}\times \frac{\sqrt{g}}{10(\sqrt{2}-1)}=\frac{1}{0.4}=\frac{10}{4}=\frac{5}{2}\)

3. The ratio of times to reach the ground and to reach first half of the distance is

\(t:{{t}_{1}}=\frac{10\sqrt{2}}{\sqrt{g}}\times \frac{\sqrt{g}}{10}=\sqrt{2}\)

Consider an inclined plane which is making an angle \(\beta\) with the horizontal and a particle is projected from point O with a velocity u m s^–1 making an angle \(\alpha \) with the horizontal, and strike at point B on the inclined plane as shown in figure. The angle of projection is \((\alpha -\beta )\) from incline plane. Then, assume OB along x-axis and y–axis is perpendicular to OB . Then according to vector rule initial velocity will be

Acceleration vector \(\vec{a}=-\,\,\vec{g}\sin \beta \ \hat{i}-\vec{g}\cos \beta \ \hat{j}\)

Using \(\vec{y}=\vec{u}\sin (\alpha -\beta )t-\frac{1}{2}\vec{g}\cos \beta \,{{t}^{2}}\)       …(11)

and \(\vec{x}=\vec{u}\cos \,(\alpha -\beta )\,t-\frac{1}{2}\vec{g}\sin \beta \,{{t}^{2}}\)         …(12)

Time of flight:

Putting y = 0 in equation (11) (when particle strike at point B its y coordinate is zero) we get

\(T=\frac{2\left| {\vec{u}} \right|\sin (\alpha -\beta )}{|\vec{g}|\cos \beta }\)                             …(13)

Horizontal range:

Horizontal displacement,\((\overrightarrow{OC})=\vec{u}\cos \alpha T\)

\(\therefore \)  \(\overrightarrow{OC}=2\vec{u}\cos \alpha \frac{|\vec{u}|\sin (\alpha -\beta )}{|\vec{g}|\cos \beta }\)

\(\overrightarrow{OB}=\overrightarrow{R}=\frac{\overrightarrow{OC}}{\cos \beta }=\frac{2\vec{u}\cos \alpha |\vec{u}|\sin (\alpha -\beta )}{|\vec{g}|{{\cos }^{2}}\beta }\)

or \(|\overrightarrow{R}|\ =\frac{|\vec{u}{{|}^{2}}[\sin (2\alpha -\beta )-\sin \beta ]}{|\vec{g}|{{\cos }^{2}}\beta }\)

For Maximum Range

\(\sin (2\alpha -\beta )=1\)

\((2\alpha -\beta )=\frac{\pi }{2}\) or \(\alpha =\left( \frac{\pi }{2}+\frac{\beta }{2} \right)\)

Thus, \(|{{\overrightarrow{R}}_{\max }}|\ \ =\frac{|\vec{u}{{|}^{2}}(1-sin\beta )}{|\vec{g}|{{\cos }^{2}}\beta }=\frac{|\vec{u}{{|}^{2}}}{|\vec{g}|(1+\sin \beta )}\)

If particle is projected down the plane, \(\beta \) will be replaced  to – \(\beta\)

So,\(|{{\overrightarrow{R}}_{\max }}|\ \ =\frac{|\vec{u}{{|}^{2}}}{|\vec{g}|(1-\sin \beta )}\)

Maximum height:

From equation (11) and (13)

\(\vec{y}=\vec{u}\sin (\alpha -\beta )t-\frac{1}{2}\vec{g}\cos \beta \,{{t}^{2}}\)  and \(T=\frac{2\left| {\vec{u}} \right|\sin (\alpha -\beta )}{|\vec{g}|\cos \beta }\)

For maximum height use \(t=\frac{T}{2}\) in equation (11)

\(\vec{H}=\vec{u}\sin (\alpha -\beta )t-\frac{1}{2}\vec{g}\cos \beta \,{{t}^{2}}\) on solving we get,

\(\vec{H}=\frac{|\vec{u}{{|}^{2}}{{\sin }^{2}}(\alpha -\beta )}{2|\vec{g}|\cos \beta }\)

1. Time taken by the projectile to reach from A to B is t. Then the distance AB is equal to

Solution: 

Time taken by the particle to go from A to N ( along  horizontal ) is also t

In \(\Delta ABN\)

\(\cos 30=\frac{AN}{AB}\)

\(\Rightarrow AB=\frac{AN}{\cos 30}=\frac{ut\cos \left( {{60}^{0}} \right)}{\cos \left( {{30}^{0}} \right)}\)

\(\Rightarrow AB=\frac{ut}{\sqrt{3}}\)

2. A ball starts falling freely from a height \[h\] from a point on the inclined plane forming an angle \(\alpha\) with the horizontal as shown. After collision with the incline it rebounds elastically off the inclined plane.Then

Solution: 

\(0=\sqrt{2gh}\left( \cos \alpha \right)t-\frac{1}{2}\left( g\cos \alpha \right){{t}^{2}}\)

\(\Rightarrow t=\sqrt{\frac{8h}{g}}\) after  it strikes the incline at A

\(s=\sqrt{2gh}\left( \sin \alpha \right)t+\frac{1}{2}\left( g\sin \alpha \right){{t}^{2}}\)

Substituting  value of t,we get = s = 8h sin\(\alpha \)

1. A particle is projected from ground. The co-ordinate axes are shown in the diagram. The minimum velocity with which projectile should be thrown so that it passes through a point (30m, 40m) is \(10x\text{ }m{{s}^{-1}}\).  [g=10 ms^-2] find \(x\)

Solution: 

\(\begin{align} & {{R}_{\max }}=\frac{{{u}^{2}}}{g\left( 1+\sin \alpha \right)} \\ & {{u}_{\min }}=\sqrt{g\left( 1+\sin \alpha \right)R} \\ & =\sqrt{10\left( 1+\frac{4}{5} \right)\times 50} \\ & =\sqrt{10\times \frac{9}{5}\times 50} \\ & =\sqrt{900}m{{s}^{-1}} \\ & =30m{{s}^{-1}} \\ \end{align}\)

2. A particle is projected with a velocity of 10 ms^-1 at an angle 37⁰ with horizontal. A bird moves along the path of projected particle with constant speed of 4 ms^-1. Find the acceleration of bird when it is at highest point of path of particle.(g = 10ms^-2)

Solution: 

For particle \(Mg=\frac{M{{\left( u\cos \theta \right)}^{2}}}{R}\)

\(\Rightarrow R=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{g}\)

For Bird, \({{a}_{n}}=\frac{{{v}^{2}}}{R}=\frac{{{v}^{2}}}{{{u}^{2}}{{\cos }^{2}}\theta }\times g\)

=\(\frac{{{4}^{2}}}{{{10}^{2}}{{\cos }^{2}}{{37}^{0}}}\times 10\)

= 2.5 ms^-2

1. The front windscreen of a car is inclined at an angle \(30{}^\text{o}\) with the horizontal. Hailstones fall vertically downwards with a speed of \(5\sqrt{3}\) m/s. The speed of the car so that hailstones are bounced back by the screen in vertically upward direction with respect to car is (Assume elastic collision of hailstones with wind screen)

Solution: 

\(\tan 60{}^\text{o}=\frac{{{v}_{C}}}{{{v}_{H}}}\)

\(\therefore {{v}_{C}}={{v}_{H}}\,\tan 60{}^\text{o}=15m/s\)

2. A particle is projected from point A at an angle of \(53{}^\text{o}\)with horizontal. At the same time wedge starts from rest and moves with constant acceleration a as shown. The value of a for which the particle strikes the plane perpendicular to it at \(P\,\,is\,\,\left( \tan 53{}^\text{o}=\frac{4}{3} \right)\left( g=10m/{{s}^{2}} \right)\)

Let the height of point P above the ground be h.

\(u\,\cos 53{}^\text{o}\,t+h=\frac{1}{2}a{{t}^{2}}\)

\(\frac{u\,\,\sin 53{}^\text{o}-gt}{u\,\,\cos 53{}^\text{o}}=1\)

\(h=u\,\,\sin 53{}^\text{o}t-\frac{1}{2}g{{t}^{2}}\)

1.  A particle is projected from a point at the foot of a fixed plane inclined at an angle of \({{45}^{0}}\) to the horizontal in thevertical plane containing the line of greatest slope through the point. If the particle strikes the plane horizontally and \(\phi \left( >{{45}^{0}} \right)\) is the angle of launch measured to the horizontal then find  the value of tan\(\phi\).

Solution: 

Let the particle be projected from O with velocity u and strike the

plane at a point P horizontally . Then

PQ = OQ

\(\Rightarrow\) maximum height =\(\frac{Horizontal\,range}{2}\)

\(\Rightarrow \frac{{{u}^{2}}{{\sin }^{2}}\phi }{2g}=\frac{{{u}^{2}}\sin 2\phi }{2g}=\frac{{{u}^{2}}\sin \phi \cos \phi }{g}\)

\(\Rightarrow \tan \phi =2\)

2. A particle is projected from a point at the  foot of  a fixed plane inclined at an angle of \({{45}^{0}}\) to the  horizontal  in the vertical plane containing the line of greatest slope through the point. If the particle strikes  the plane at right angles and \(\phi \left( >{{45}^{0}} \right)\) is the angle of launch measured to the horizontal then find  the value of tan\(\phi\).

Solution: 

At time t = T = \(\frac{2u\sin \left( \phi -{{45}^{0}} \right)}{g\cos {{45}^{0}}}\)

Component of velocity  along  the plane is zero

\(\Rightarrow 0=u\cos \left( \phi -{{45}^{0}} \right)-\left( g\sin {{45}^{0}} \right)t\)

\(\Rightarrow u\cos \left( \phi -{{45}^{0}} \right)=\left( g\sin {{45}^{0}} \right)\left( \frac{2u\sin \left( \phi -{{45}^{0}} \right)}{g\cos {{45}^{0}}} \right)\)

\(\Rightarrow 2\tan \left( \phi -{{45}^{0}} \right)=\cot {{45}^{0}}=1\)

\(\Rightarrow 2\left( \frac{\tan \phi -\tan {{45}^{0}}}{1+\tan \phi \tan {{45}^{0}}} \right)=1\)

\(\Rightarrow \tan \phi =3\)