Question

A stone is thrown upwards from the top of a tower with some initial speed and it reaches the ground in `16` seconds. Now it is thrown with the same initial speed downward and it reaches the ground in `9 sec`. In how much time will it reach the ground if the stone is allowed to fall freely under gravity from the same place ?
A. 25 s
B. 12.5 s
C. 7 s
D. 12 s

Answer 1

Correct Answer - D
Suppose the body is projected vertically upwards from `A` with a speed `u_0`.
Using equation `s = ut + ((1)/(2)) at^2`
For first case, `- h = u_0 t_1 - ((1)/(2))g t_1^2` ….(i)
For second case, `A - h = - u_0 t_2 - ((1)/(2)) g t_2^2` …(ii)
`(i) - (ii) 0 = u_0 (t_2 + t_1) + ((1)/(2))g (t_2^2 - t_1^2)`
`u_0 = ((1)/(2)) g(t_1 - t_2)` ...(iii)
Put the value of `u_0` in (ii),
`-h = -((1)/(2))g (t_1 - t_2) t_2 - ((1)/(2)) g t_2^2`
`rArr h = (1)/(2) g t_1 t_2` ...(iv)
For third case, `u = 0, t = ?`
`-h = 0 xx t - ((1)/(2)) g t^2` or `h = ((1)/(2))g t^2`...(v)
Combining equation (iv) and equation (v), we get
`(1)/(2) g t^2 = (1)/(2) g t_1 t_2` or `t = sqrt(t_1 t_2)`
Putting the value we get `t = 12 sec`.