Let a particle is projected from the ground with velocity u m s–1\(\theta\) with the ground and it follow a curved path OAB as show in figure. at an angle
Then according to vector rule
\(\vec{u}=u\cos \theta \ \hat{i}+u\sin \theta \ \hat{j}\)
As velocity along x-axis \({{\vec{u}}_{x}}=\vec{u}\cos \theta \) and velocity along y-axis \({{\vec{u}}_{y}}=\vec{u}\sin \theta\)
Using \({{\vec{v}}_{y}}={{\vec{u}}_{y}}+{{\vec{a}}_{y}}t\ \) (where \({{\vec{a}}_{y}}\) is the acceleration of the object in the vertical direction = –g)
\({{\vec{v}}_{y}}=\vec{u}\sin \theta -\vec{g}\ t\) ….. (1)
and \({{\vec{v}}_{x}}={{\vec{u}}_{x}}+{{\vec{a}}_{x}}t\)
\({{\vec{v}}_{x}}=u\cos \theta +0.\ t\) (where \({{\vec{a}}_{x}}\) is the acceleration along horizontal direction = 0)
\({{\vec{v}}_{x}}=u\cos \theta\) ….. (2)
Thus velocity at any instant
\(\vec{v}=u\cos \theta \ \hat{i}+(u\sin \theta -gt)\hat{j}\) ….. (3)
Similarly, displacement along y-axis
\(\vec{y}=\vec{u}\sin \theta \ t-\frac{1}{2}\vec{g}{{t}^{2}}\) ….. (4)
and along x axis will be
\(\vec{x}=\vec{u}\cos \theta \ t\) ….. (5)
Note : Path of projection is parabolic :
From relation (4) and (5) eliminating t, we get
\(y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\) ….. (6)
If we take A = \(\tan \theta\)
and \(B=\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }\)
So, \(y=Ax-B{{x}^{2}}\)
This equation represent that path of projection is parabola.
1.A particle is moving in x-y plane. At certain instant of time the components of its velocity and acceleration are \({{v}_{x}}=3m{{s}^{-1}}\), \({{v}_{y}}=4m{{s}^{-1}}\), \({{a}_{x}}=2m{{s}^{-2}}\) and \({{a}_{y}}=1m{{s}^{-2}}\) ,The rate of change of speed at this moment is
Solution:
Since \({{v}^{2}}={{v}_{x}}^{2}+{{v}_{y}}^{2}\)
\(\Rightarrow 2v\frac{dv}{dt}=2{{v}_{x}}\frac{d{{v}_{x}}}{dt}+2{{v}_{y}}\frac{d{{v}_{y}}}{dt}\)
\(\Rightarrow \frac{dv}{dt}=\frac{{{v}_{x}}{{a}_{x}}+{{v}_{y}}{{a}_{y}}}{v}=\frac{{{v}_{x}}{{a}_{x}}+{{v}_{y}}{{a}_{y}}}{\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}}\)
\(\Rightarrow \frac{dv}{dt}=\frac{\left( 3 \right)\left( 2 \right)+\left( 4 \right)\left( 1 \right)}{5}=2m{{s}^{-1}}\)
2. A rod of length \(l\) leans by its upper end against a smooth vertical wall, while its other rend leans against the floor. The end that leans against the wall moves uniformly downward with velocity \({{v}_{0}}\). Then the situation discussed is shown at time t . Then we observe that at any instant
Solution:
\({{l}^{2}}={{x}^{2}}+{{y}^{2}}\)
\(\Rightarrow 0=2x\frac{dx}{dt}+2y\frac{dy}{dt}\)
\(\Rightarrow \frac{dx}{dt}=-\frac{y}{x}\left( \frac{dy}{dt} \right)\)
From ( 1) \(x=\sqrt{{{l}^{2}}-{{y}^{2}}}\)
\(\Rightarrow \frac{dx}{dt}=\frac{y{{v}_{0}}}{\sqrt{{{l}^{2}}-{{y}^{2}}}}\)
So the speed of the lower end decreases and vanishes when \(y\to 0\)
1. A jet plane flying at a constant velocity \(v\) at a height \(h=8\) kilometer is being tracked by a radar R located at O directly below the line of flight. If the angle \(\theta\) is decreasing at the rate of 0.025 rads-1 the velocity of the plane when \(\theta =60{}^\circ\) is
Solution:
\({{h}^{2}}={{r}^{2}}-{{x}^{2}}\)
\(\tan \theta =\frac{h}{x}\)
\(h=r\sin \theta\)
\(\Rightarrow x=h\cot \theta\)
\(\Rightarrow x=h\cot \theta\)
\(\Rightarrow \frac{dx}{dt}=-\cos e{{c}^{2}}\theta \left( \frac{d\theta }{dt} \right)\)
\(\Rightarrow \frac{dx}{dt}=-\left( 8000 \right)\cos e{{c}^{2}}\left( 60 \right)\left( -0.025 \right)\)
\(\Rightarrow \frac{dx}{dt}=\left( 8000 \right)\left( \frac{4}{3} \right)\left( 0.025 \right)\)
\(\Rightarrow \frac{dx}{dt}=\frac{800}{3}m{{s}^{-1}}\)
\(\Rightarrow \frac{dx}{dt}=\frac{800}{3}\times \frac{18}{5}\)
\(\Rightarrow v=960km{{h}^{-1}}\)
2. A ball is thrown from the top of a staircase which just touches the ceiling and finally hits the bottom of the steps.The initial speed of the ball is
Solution:
\(4-\left( u\cos \alpha \right)t\)
\(2=\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g}\)
\(-3=\left( u\sin \alpha \right)t-\frac{1}{2}g{{t}^{2}}\)
From (2) \(u\sin \alpha =2\sqrt{g}\)
Substituting in ( 3) we get
\(-3=6.3t-4.9{{t}^{2}}\)
\(\Rightarrow 4.9{{t}^{2}}-6.3t-3=0\)
\(\Rightarrow t=\frac{6.3\pm \sqrt{39.7-4\left( 4.9 \right)\left( -3 \right)}}{2\left( 4.9 \right)}\)
\(\Rightarrow t=\frac{6.3\pm \sqrt{98.5}}{9.8}\)
\(\Rightarrow t=\frac{6.3\pm 9.9}{9.8}=8.1s\)
From (1) 4 =\(\left( u\cos \alpha \right)\left( 8.1 \right)\)
\(\Rightarrow u\cos \alpha =\frac{4}{8}\cong \frac{1}{2}\)
From ( 4) and ( 5 ) we get
\({{u}^{2}}{{\sin }^{2}}\alpha +{{u}^{2}}{{\cos }^{2}}\alpha =40+\frac{1}{4}\\\Rightarrow u=\sqrt{40.25}=6.3m{{s}^{-1}}\)
3. A particle moves in the x-y plane with constant acceleration a directed along the negative y-axis .The equation of motion of the particle has the form \(y=\alpha x-\beta {{x}^{2}}\), where \(\alpha\) and \(\beta\)are positive constants . The velocity of the particle at the origin of coordinates is \(y=\alpha x-\beta {{x}^{2}}\)
Solution:
Comparing with standard equation of projectile
\(y=x\tan \theta -\frac{a{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\) we get
\(\alpha =\tan \theta \, and \,\beta =\frac{a}{2{{u}^{2}}}{{\sec }^{2}}\theta \)
Since \({{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta\)
\(\therefore \beta =\frac{a}{2{{u}^{2}}}\left( 1+{{\alpha }^{2}} \right)\)
\(\Rightarrow u=\sqrt{\frac{a}{2\beta }\left( 1+{{\alpha }^{2}} \right)}\)
1. A large number of bullets are fired in all the directions with the same speed \(v\). The maximum area on the ground on which these bullets will spread is
Solution:
\({{R}_{\max }}=\frac{{{v}^{2}}}{g}at\theta ={{45}^{0}}\)
Maximum Area = \(\pi {{\left( {{R}_{\max }} \right)}^{2}}=\frac{\pi {{v}^{4}}}{{{g}^{2}}}\)
\(R=\left( a+b \right)=\frac{{{u}^{2}}\sin \left( 2\alpha \right)}{g}=\frac{{{u}^{2}}}{g}\)
\(\Rightarrow {{u}^{2}}=g\left( a+b \right)\)
Since y = a\(\tan \alpha\)- \(\frac{g{{a}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\alpha }\)
\(\Rightarrow h=a\tan \alpha -\frac{g{{a}^{2}}}{2g\left( a+b \right){{\cos }^{2}}\alpha }\)
\(\Rightarrow h=a\tan 45-\frac{g{{a}^{2}}}{2g\left( a+b \right){{\cos }^{2}}45}\)
\(\Rightarrow h=\frac{ab}{a+b}\)
2. A particle is projected with a certain velocity at an angle \(\alpha \) above the horizontal from the foot of an inclined plane of inclination \({{30}^{0}}\). If the particle strikes the plane normally then \(\alpha\) is equal to
Solution:
If T be projectile to hit the plane normally \({{v}_{x}}=0\)
\(\Rightarrow {{u}_{x}}+{{a}_{x}}T=0\)
\(\Rightarrow u\cos \left( \alpha -{{30}^{0}} \right)-g\sin \left( {{30}^{0}} \right)T=0\)
\(\Rightarrow u\cos \left( \alpha -{{30}^{0}} \right)-g\sin \left( {{30}^{0}} \right)\left( \frac{2u\sin \left( \alpha -{{30}^{0}} \right)}{g\cos \left( {{30}^{0}} \right)} \right)\)
\(\Rightarrow \tan \left( \alpha -{{30}^{0}} \right)=\frac{1}{2}\cot \left( {{30}^{0}} \right)\)
\(\Rightarrow \tan \left( \alpha -{{30}^{0}} \right)=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \alpha ={{30}^{0}}+{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\)
3. A body of mass m thrown horizontal with velocity \(v\) from the top of the tower of height h touches the ground at a distance of 250m from the foot of the tower. A body of mass 2m thrown with a velocity of \(\frac{v}{2}\) from the top of the tower of height 4h will touch the ground at a distant of
Solution:
For a horizontal projectile we have
\(y=\frac{g{{x}^{2}}}{2{{u}^{2}}}\) So
\(h=\frac{g{{\left( 250 \right)}^{2}}}{2{{v}^{2}}}\)
Similarly
4h =\(\frac{g{{x}^{2}}}{2{{\left( v/2 \right)}^{2}}}\)
Dividing (1) and ( 2) , we get X=250 m
1. A boy throws a ball upwards with velocity \({{v}_{0}}=20m{{s}^{-1}}\).The wind imparts a horizontal acceleration of 4 \(m{{s}^{-2}}\) to the left.The angle \(\theta\) at which the ball must be thrown so that the ball returns to the boy’s hands is \(\left( g=10m{{s}^{-2}} \right)\)
Solution:
If T be the time of flight of the ball then
\(T=\frac{2{{V}_{y}}}{g}=\frac{2\left( {{v}_{0}}\cos \theta \right)}{g}\)
\(\Rightarrow T=\frac{2\left( 20\cos \theta \right)}{10}=4\cos \theta\)
For the ball to return to the boy’s hand we have x= 0
\(\Rightarrow 0={{v}_{x}}T+\frac{1}{2}{{a}_{x}}{{T}^{2}}\)
\(\Rightarrow 0=\left( {{V}_{0}}\sin \theta \right)T+\frac{1}{2}\left( -4 \right){{T}^{2}}\)
\(\Rightarrow 0=20\sin \theta -2T\)
\(\Rightarrow T=10\sin \theta\)
From (1) and (2) we get
\(4\cos \theta =10\sin \theta\)
\(\Rightarrow \cot \theta =2.5\)
\(\theta ={{\cot }^{-1}}\left( 2.5 \right)\)
2. A number of projectiles each with a fixed muzzle velocity u are fired at different angles lying between \(0{}^\circ\) and \(90{}^\circ \) as shown. Neglecting are resistance and assuming g to be constant, then the equation of the envelope E of the parabolic trajectories (as shown in figure) is
Solution:
\(y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\)
\(y=mx-\frac{g{{x}^{2}}}{2{{u}^{2}}}\left( 1+{{m}^{2}} \right)\,\,\,where\,\,m=\tan \theta\)
For \(\theta\) to be real we must have
\(\Rightarrow \left( \frac{g{{x}^{2}}}{2{{u}^{2}}} \right){{m}^{2}}-xm+\left( \frac{g{{x}^{2}}}{2{{u}^{2}}}+y \right)=0\)
Equation (1) is a quadratic in m with two roots \({{m}_{1}}\) and \({{m}_{2}}\).These two roots \({{m}_{1}}\) and \({{m}_{2}}\) of equation ( 1) give up the two firing angles for the two trajectories shown ( in the statement ) such that the projectiles pass through a common point A.This point A will approach the envelope E as the two roots approach equality i.e.the DISCRIMINANT for quadratic in m must equal zero
\({{x}^{2}}-4\left( \frac{g{{x}^{2}}}{2{{u}^{2}}} \right)\left( \frac{g{{x}^{2}}}{2{{u}^{2}}}+y \right)=0\)
\(1-\left( \frac{2g}{{{u}^{2}}} \right)\left( \frac{g{{x}^{2}}}{2{{u}^{2}}}+y \right)=0\)
\(\frac{{{u}^{2}}}{2g}-\frac{g{{x}^{2}}}{2{{u}^{2}}}-y=0\)
\(\Rightarrow y=\frac{{{u}^{2}}}{2g}-\frac{g{{x}^{2}}}{2{{u}^{2}}}\)
Is the equation of the envelope E
3. The trajectory of a projectile in vertical plane is \(y=ax-b{{x}^{2}}\), a,b are constants and x,y are respectively horizontal and vertical distance of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are
Solution:
Method - I
\(y=ax-b{{x}^{2}}\)
Compare with
\(y=x\tan \theta -\left( \frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta } \right){{x}^{2}}\)
We get
\(\tan \theta =a\) and \(\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }\) = b
\({{H}_{\max }}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\left( \frac{g}{2b{{\cos }^{2}}\theta } \right)\frac{{{\sin }^{2}}\theta }{2g}\)
\(\Rightarrow {{H}_{\max }}=\frac{{{\tan }^{2}}\theta }{4b}=\frac{{{a}^{2}}}{4b}\)
And \(\theta ={{\tan }^{-1}}\left( a \right)\)
Method – II
At maximum height
\(\frac{dy}{dt}=0\)
\(\Rightarrow a-2bx=0\)
\(\Rightarrow x=\frac{a}{2b}\)
\(\Rightarrow {{y}_{\max }}=y\left| _{x=\frac{a}{2b}} \right.=a\left( \frac{a}{2b} \right)-b\left( \frac{{{a}^{2}}}{4{{b}^{2}}} \right)\)
\(\Rightarrow {{y}_{\max }}=\frac{{{a}^{2}}}{4b}\)
Now \(\frac{dy}{dx}=a-2bx\)
So the slope at the point of launch \(\left( x=0 \right)\) is tan\(\theta\). Hence
\(\frac{dy}{dx}\left| _{x=0} \right.=a=\tan \theta\)
\(\Rightarrow \theta ={{\tan }^{-1}}\left( a \right)\)
Time of flight:
Total time taken by the particle to go up and come down to the same level from where it was projected is known as time of flight denoted by T.
From equation (1)
\({{\vec{v}}_{y}}=\vec{u}\sin \theta -\vec{g}t\)
We know that \({{\vec{v}}_{y}}=0\) at height point.
So, \(0=\vec{u}\sin \theta -\vec{g}t\)
or \(t=\frac{\vec{u}\sin \theta }{{\vec{g}}}\) ….. (7)
Since time is scalar quantity. So we use only magnitude of \(\vec{u}\sin \theta\) and \(\vec{g}\). Thus time taken to reach maximum height is
\(t=\frac{|\vec{u}\sin \theta |}{|\vec{g}|}\)
and total time taken to reach the same level from where it was projected is
\(T=2t=\frac{2|\vec{u}\sin \theta |}{|\vec{g}|}\) …...(8)
Horizontal range:
It is the horizontal distance between initial and final point of projectile path. Denoted by \[\overrightarrow{R}\]
\(\overrightarrow{OB}=\vec{u}\cos \theta \ T\)
\(\overrightarrow{OB}=(\vec{u}\cos \theta )\frac{2\vec{u}\sin \theta }{{\vec{g}}}\)
So, \(\overrightarrow{OB}=\frac{|\vec{u}{{|}^{2}}\sin 2\theta }{{\vec{g}}}=\overrightarrow{R}\) ….. (9)
For maximum range,\(\sin 2\theta =1,\,\,so\,2\theta ={{90}^{o}}\,or\,\,\theta ={{45}^{o}}\)
\({{\overrightarrow{R}}_{\max }}=\frac{|\vec{u}{{|}^{2}}}{{\vec{g}}}\)
Maximum height:
Maximum vertical displacement from the ground is known as maximum height denoted by H.
From equation (4) and (7) we get,
\(H=(u\sin \theta )t-\frac{1}{2}g{{t}^{2}}\) ….. (4)
\(\left( |\vec{H}|=H,\,\,|\vec{u}|=u,\,\,|\vec{g}|=g \right)\)
And \(t=\frac{u\sin \theta }{g}\) ….. (7)
\(H=(u\sin \theta )\left( \frac{u\sin \theta }{g} \right)-\frac{1}{2}g{{\left( \frac{u\sin \theta }{g} \right)}^{2}}\)
On solving, we get
\(H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\) ….. (10)
For a particle moving in the x-y plane the x,y coordinates as a function of time are given by x=6t and \(y=8t-5{{t}^{2}}\) where x and y are in metre and \(t\) is in second. Assume no air drag answer the following questions. Based on the above facts answer the following questions
1. time of ascent of the projectile is
\({{t}_{ascent}}=\frac{{{u}_{y}}}{g}\)
\(\Rightarrow {{t}_{ascent}}=\frac{8}{10}=0.8s\)
2. The maximum height attained by the projectile is
\(H=\frac{{{u}^{2}}_{y}}{2g}\)
\(\Rightarrow H=\frac{{{\left( 8 \right)}^{2}}}{2\left( 10 \right)}\)
\(\Rightarrow H=3.2m\)
3. The horizontal range of the projectile is
\(R=\frac{2}{g}\left( {{u}_{x}} \right)\left( {{u}_{y}} \right)\)
\(\Rightarrow R=\frac{2}{10}\left( 6 \right)\left( 8 \right)\)
\(\Rightarrow R=9.6m\)
1. The height and width of each step of a staircase are \(20cm\)and \(30cm\) respectively .A ball rows off the top of a stair with horizontal velocity ‘v’ and hits the fifth step. The magnitude of v is \((g=10m{{s}^{-2}})\)
Solution:
H = 5h
H = 100 cm
H = 1 m
R = 5b
R= 150 cm
\(R=v\times \,\sqrt{\frac{2H}{g}}\)
\(\begin{align} & \frac{150}{100}=v\sqrt{\frac{2\times 1}{10}} \\ & v=1.5\sqrt{5}\,m/s \\ \end{align}\)
2. An aeroplane flying horizontally at an altitude of \(490m\) with a speed of \(180kmph\) drops a bomb. The horizontal distance at which it hits the ground is Q
Solution:
\(\begin{align} & R=V\,\times \,\,\sqrt{\frac{2H}{g}} \\ & R=50\sqrt{\frac{2\times 490}{9.8}}=500\,m \\ \end{align}\)
1. The vertical displacement y of the projectile varies with the horizontal displacement x as, \(y=ax-b{{x}^{2}}\) where a and b are constants. Then the time of flight of the projectile is
Solution:
Give equation \(y=ax-b{{x}^{2}}\)
\(\frac{dy}{dt}=a\frac{dx}{dt}-\frac{2bxdx}{dt}\)
\({{V}_{y}}=a{{V}_{x}}-2bx{{V}_{x}}\)
=\((a-2bx){{V}_{x}}\)
Mass height \({{V}_{y}}=0;\)
\(\Rightarrow x=a/2b\)
\({{h}_{\max }}=a(\frac{a}{2n})-b{{(\frac{a}{2b})}^{2}}=\frac{{{a}^{2}}}{4b}\)
Time to reach mass height
\(t=\sqrt{\frac{2{{h}_{\max }}}{g}}=\sqrt{\frac{2{{a}^{2}}}{4bg}}=\frac{a}{\sqrt{2bg}}\)
Time of flight \(2t=a\sqrt{\frac{2}{bg}}\)
2. A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of \(60{}^\circ\),with the vertical with the same initial speed. At the highest point, the ratio of their potential energies will be
Solution:
Max height \({{h}_{1}}=\frac{{{U}^{2}}}{2g}\)
U is the initial speed maximum height attained by the second ball is (\(\theta =90{}^\circ -60{}^\circ =30{}^\circ\))
\({{h}_{2}}=\frac{{{U}^{2}}{{\sin }^{2}}30}{2g}=\frac{{{U}^{2}}}{2g}\)
P.E ball \(1\to {{h}_{1}}=mg{{h}_{1}}\) Ratio \(=\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{{{U}^{2}}}{2g}\times \frac{8g}{{{U}^{2}}}\)
Ball \(2\to {{h}_{2}}=mg{{h}_{2}}\)
3. A shell is fired from a gun from the bottom of hill along its slope.The slope of the hill is \({{30}^{0}}\) and the angle of the barrel to the horizontal \({{60}^{0}}\). The initial velocity of the shell is \(21m{{s}^{-1}}\). Find the distance inmetre from the gun to the point at which the shell falls.
Solution:
Since \(R=\frac{2{{u}^{2}}\sin \left( \alpha -\beta \right)\cos \alpha }{g{{\cos }^{2}}\beta }\)
\(\Rightarrow R=\frac{2{{\left( 21 \right)}^{2}}\sin \left( {{60}^{0}}-{{30}^{0}} \right)\cos \left( {{60}^{0}} \right)}{9.8{{\cos }^{2}}\left( {{30}^{0}} \right)}\)
\(\Rightarrow R=\frac{\left( 2 \right)\left( 441 \right)\left( \frac{1}{2} \right)\left( \frac{1}{2} \right)}{\left( 9.8 \right)\left( \frac{3}{4} \right)}=30m\)
1. An oblique projectile islaunched from a point O with an initial velocity \(u\) that makes an angle \(\theta\) with the horizontal . It is observed that the projectile attains a maximum height H at the point \(\left( \frac{R}{2},H \right)\), where R is the maximum distance from O, at which the projectile strikes the ground. Now if the velocity of the projectile is \(\overrightarrow{v}=\left( 20\hat{i}+10\hat{j} \right)m{{s}^{-1}}\) when it is at a height of 15m above the ground then based on this information and taking \(g=10m{{s}^{-2}}\)
1. The launch speed in is
Since \(\overrightarrow{v}=20\hat{j}+10\hat{j}\)
\(\Rightarrow {{v}_{x}}={{u}_{x}}=20m{{s}^{-1}}\)
Since \({{v}_{y}}=10m{{s}^{-1}}\, and \,{{v}_{y}}^{2}-{{u}^{2}}_{y}=2{{a}_{y}}y\)
\(\Rightarrow {{\left( 10 \right)}^{2}}-{{u}^{2}}_{y}=2\left( -10 \right)\left( 15 \right)\)
\(\Rightarrow {{u}^{2}}_{y}=400\)
\(\Rightarrow {{u}_{y}}=20m{{s}^{-1}}\)
So \(u=\sqrt{{{u}_{x}}^{2}+{{u}^{2}}_{y}}=20\sqrt{2}m{{s}^{-1}}\)
2.The value of \(\theta\)( in radian ) is
Since \(\tan \theta =\frac{{{u}_{y}}}{{{u}_{x}}}=\frac{20}{20}=1\)
\(\Rightarrow \theta ={{45}^{0}}=\frac{\pi }{4}\) radian
3. The coordinate where the maximum height is attained is
\(H=\frac{{{u}^{2}}_{y}}{2g}=\frac{400}{20}=20m\)
\(R=\frac{{{u}^{2}}\sin \left( 2\theta \right)}{g}=\frac{{{\left( 20\sqrt{2} \right)}^{2}}\sin \left( 90 \right)}{10}\)
\(\Rightarrow R=80m\)
Now the coordinate where the maximum height is attained is
\(\left( \frac{R}{2},H \right), So \,\,we\,\, get \,\,\left( 40,20 \right)\) m as the answer
2. A projectile is thrown with an initial velocity \(u\) at an angle of projection \(\theta\) first from the equator and then from the pole.The fractional decrement in the range of projectile is
Solution:
Since R =\(\frac{{{u}^{2}}\sin \left( 2\theta \right)}{g}\)
\(\Rightarrow R=k{{g}^{-1}}\)
\(\Rightarrow dR=-k{{g}^{-2}}dg\)
\(\Rightarrow \frac{dR}{R}=\frac{-k{{g}^{-2}}}{k{{g}^{-1}}}\)
\(\Rightarrow \frac{dR}{R}=-\frac{dg}{g}=-\left( \frac{{{g}_{p}}-{{g}_{e}}}{g} \right)\)
\(\Rightarrow \) Fractional Decrement =\(\frac{dR}{R}=\frac{1}{291}\)
Also \({{R}_{e}}=\) Range at equator =\(\frac{{{u}^{2}}\sin \left( 2\theta \right)}{{{g}_{p}}}\)
\({{R}_{P}}=\) Range at pole \(=\frac{{{u}^{2}}\sin \left( 2\theta \right)}{{{g}_{p}}}\)
\(R=\left( \begin{align} & Original\,range\,with \\ & no\,\operatorname{var}iation\,of\,g \\ & taken\,\operatorname{int}o\,account \\ \end{align} \right)=\frac{{{u}^{2}}\sin \left( 2\theta \right)}{g}\)
\(\Rightarrow \frac{dR}{R}=\left( \begin{align} & Fractional \\ & Decrement \\ \end{align} \right)=\frac{{{R}_{e}}-{{R}_{p}}}{R}\)
\(\Rightarrow \frac{dR}{R}=\frac{{{u}^{2}}\sin \left( 2\theta \right)\left( \frac{1}{{{g}_{e}}}-\frac{1}{{{g}_{p}}} \right)}{{{u}^{2}}\sin \left( 2\theta \right)\frac{1}{g}}\)
\(\Rightarrow \frac{dR}{R}=-g\left( \frac{1}{{{g}_{p}}}-\frac{1}{{{g}_{e}}} \right)\)