Projection from Top of Tower

Projection from Top of Tower

PROJECTION FROM TOP OF TOWER

(EXPRESSION FOR HEIGHT OF TOWER):
Consider a tower of height H. Suppose a body is projected upwards vertically with initial velocity from the top of tower. Suppose it reaches to a displacement x above the tower and there after reaches the foot of the tower. Let t be the total time of travel.
Now, considering the total path of the body,the motion parameters are as follows.
    Initial velocity of the body = u
    Net displacement of body = S =+x-x-H =-H 
    Time of travel = t
    we know that, \( S = ut + \frac{1} {2}at^2 \)
                

    Here,                                    
    a=-g, SH, u = u, t = t
    \( \therefore - H = ut - \frac{1} {2}gt^2 \)
    \( \therefore H = \frac{1} {2}gt^2 - ut \)

Projection from Top of Tower

Application-1:

A balloon is rising up vertically. A stone is dropped from the balloon, when the velocity bal- loon is 'u'. At the instant, the stone velocity is also 'u' vertically upwards. So the stone moves upwards until its velocity becomes zero and again fall down to ground in 't' sec, just like a body projected from the top of a tower.
                               

The height of the balloon from the ground when stone is dropped, is
                     \( h = - ut + \frac{1} {2}gt^2 \)
The stone is a freely falling body with respect to balloon.
The height of balloon from the ground when the stone reaches the ground is  \( h^1 = \frac{1} {2}gt^2 \)  
 If the balloon is rising with upward acceleration 'a' the distance between balloon and stone after 't' sec is  s=\( \frac{1} {2}(g + a)t^2 \)

Projection from Top of 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 1st body (t₁-t₂) is equal to the time of flight of 1st 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= \( \frac{1} {2}t_1t_2\)
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,  \( h = \frac{1} {2}gt_1 t_2 \) \( \Rightarrow t_1 t_2 = \frac{{2h}} {g} = t^2 \)
         \( t_{free} = \sqrt {\frac{{2h}} {g}} = \sqrt {t_1 t_2 } \)   

Projection from Top of Tower

Application-3 :

A body projected vertically upwards from ground is at the same height h from the ground at two instants of time t₁ and t₂ (both being measured from the instant of projection) Now 
a)   \( h = \frac{1} {2}gt_1 t_2 \)
b) Velocity of projection = \( u = \frac{1} {2}g(t_1 + t_2 ) \)
c)  \( H_{\max } = \frac{1} {8}g(t_1 + t_2 )^2 \)
d)A body dropped from height h takes time \( \sqrt {t_1 t_2 } \) to reach the ground