Horizontal Projectile

Application-1 :
If a body projected horizontally with velocity u from the top of a tower strikes the ground at an angle of 45°
        
    
           \( \begin{gathered} v_y = v_x ,gt = u \hfill \\ \therefore t = \frac{u} {g} \hfill \\ \end{gathered} \)

Application-2
A body is projected horizontally from the top of a tower. The line joining the point of projection and the striking point make an angle of 45° with the ground. Then 
h = u.
                         
              
                    \( \begin{gathered} \frac{1} {2}gt^2 = ut \hfill \\ t = \frac{{2u}} {g} \hfill \\ \end{gathered} \)

Horizontal Projectile

Application-3:
A body is projected horizontally from the top of a tower. The line joining the point of projection and the striking point make an angle of 45° with the ground. Then, the displacement=
       \(\sqrt 2h\)  or  \(\sqrt 2x\)
                 

From the figure tan 45° =1=\( \frac{h} {X} \Rightarrow h = X \)
.. displacement AC =\( \sqrt {\left( {AB} \right)^2 + \left( {BC} \right)^2 } \)
    = \( \sqrt {h^2 + H^2 } = \sqrt 2 h \)   (or)
     \( \sqrt 2 X \) (since h = X)

Application-4:
Two towers having heights  and  are separated by a distance 'd'. A person throws a ball horizontally with a velocity u from the top of the 1^st  tower to the top  
                               

of the 2nd tower, then  Time taken,t=\( \sqrt {\frac{{2(h_1 - h_2 )}} {g}} \)
Distance between the towers
    d = ut=u\( \sqrt {\frac{{2(h_1 - h_2 )}} {g}} \)   

Horizontal Projectile

Application-5
An aeroplane flies horizontally with a velocity 'u'. If a bomb is dropped by the pilot when the plane is at a height 'h' then
a) the path of such as body is a vertical straight line as seen by the pilot and
b) The path is a parabola as seen by an observer on the ground
c) the body will strike the ground at a certain horizontal distance. This distance is equal to the range given by
    x = ut = u\( \sqrt {\frac{{2u}} {g}} \)

Application-6
A ball rolls off the top of a staircase with a horizontal velocity u. If each step has height h and width b, then the ball will just hit the nth step if n equals to
                     

sol: nb=ut and nh=\( \frac{1} {2}gt^2 \Rightarrow n = \frac{{2hu^2 }} {{gb^2 }} \)  

Application-7
From the top of a tower, one stone is thrown towards east with velocity u, and another is thrown towards north with velocity u,. The distance be- tween then after striking the ground.
                                    \( d = t\sqrt {u_1^2 + u_2^2 } \)

Horizontal Projectile

Application-8:
Two bodies are thrown horizontally with velocities u₁,u₂  in mutually opposite directions from the same height. (or)
A bomb at rest explodes into two fragments at a height 'h' and moves horizontally in opposite di- rections with velocities u₁ and u₂ respectively. Then
a) time after which velocity vectors are perpendicular is
                \( t = \frac{{\sqrt {u_1 u_2 } }} {g} \)
For velocity vectors to be perpendicular after a time t, their dot product must be zero.
           \( \therefore \overline {v_1 } .\overline {v_2 } = 0 \) 
           \( \left( {u_1 \hat{i} - gt\hat{j}} \right).( - u_2 \hat{i} - gt\hat{j}) = 0 \)
            \( \therefore t = \frac{{\sqrt {u_1 u_2 } }} {g} \)
    
b) Separation between them when velocity vectors are perpendicular is
        X=(\( u_1 + u_2 \))\( t = \frac{{(u_1 + u_2 )\sqrt {u_1 u_2 } }} {g} \)
c) Time after which their displacement vectors are perpendicular is \( t = \frac{{2\sqrt {u_1 u_2 } }} {g} \)