Question

A body is projected horizontally from the top of a building of height h, velocity of projection is u. Find:
1. The time of will take to reach the ground.
2. Velocity with which the body reaches the ground.

Answer 1

h= 1/2× g×t^2
t= root(2h/g).
v net= root(u^2 + 2gh)

Answer 2

"First let us draw a rough diagram of the body.

Let the horizontal component of the velocity $u_x be u_x$

Let the vertical component of the velocity $u_y be 0$

Horizontal component of distance = $S_x = r$

Vertical component of distance $S_y= h$

Acceleration on horizontal range $a_x= 0$

Acceleration on vertical range $a_y= g$

Along the vertical component $S_y= u_y + \frac {1}{2}a_yt^2$

$h = 0 + \frac {1}{2}gt^2$

$h = \frac {1}{2}gt^2$

time taken for the fall $t=\sqrt { \frac { 2h }{ g } }$

Formula for range r or distance = $speed \times time\quad taken =u\times \sqrt { \frac { 2h }{ g } }$

$r =u\times \sqrt { \frac { 2h }{ g } }$

Hence $u=r\sqrt { \frac { g }{ 2h } }$"