Question

A stone thrown from the ground just crosses a wall of height 5m in 2 seconds. If the wall is at a horizontal distance of 25m from the point of projection of the stone, find at what distance the stone falls behind the wall.

Answer 1

horizontal speed of the wall from point of projection is given as

$v_x = \frac{displacement}{time}$

$v_x = \frac{25}{2}$

$v_x = 12.5 m/s$

now for vertical direction

$y = v_y*t + \frac{1}{2}at^2$

$5 = v_y* 2 + \frac{1}{2}*(-10)*2^2$

$vy = 12.5 m/s$

Now the time of flight of the projectile motion is

$T = \frac{2v_y}{g}$

$T = \frac{2*12.5}{10}$

$T = 2.5 s$

now the range is given as

$R = v_x * T$

$R = 12.5 * 2.5 = 31.25 m$

Now the distance from the wall where ball will land

$d = 31.25 - 25 = 6.25 m$

so the distance is 6.25 m from the wall