Projection From Top Of A Tower

Application-2:

Three bodies are projected from towers of same height as shown. 1st one is projected vertically up with a velocity 'u'. The second one is thrown down vertically with the same velocity and the third one is dropped as a freely falling body. If t,, t, t, are the times taken by them to reach ground, then,

a) velocity of projection is \( u = \frac{1} {2}g(t_1 - t_2 ) \)  
                
 

Sol. Clearly the extra time taken by the 1^st  body (t₁-t₂) is equal to the time of flight of 1^st  body above the tower.
i.e.,   t₁-t₂ = \( \frac{{2u}} {g} \Rightarrow u = \frac{1} {2}g(t_1 - t_2 ) \)

b)height of the tower is h=\( \frac{1} {2}t_1 t_2 \)
Sol. We know that, for a vertically projected up body \( s = ut - \frac{1} {2}gt^2 \) 

    \( \Rightarrow h = ut_1 - \frac{1} {2}gt_1 ^2 = \frac{1} {2}g(t_1 - t_2 )t_1 - \frac{1} {2}gt_1 ^2 \)  
           \( h = \frac{1} {2}gt_1 t \)

c) The time taken free fall  is given by t=\( \sqrt {t_1 t_2 } \)
    for a falling body, t=\( \sqrt {\frac{{2h}} {g}} \) 
    but,\( h = \frac{1} {2}gt_1 t_2 \) \( \Rightarrow t_1 t_2 = \frac{{2h}} {g} = t^2 \)  
          \( t_{free} = \sqrt {\frac{{2h}} {g}} = \sqrt {t_1 t_2 } \)