Question

A stone is thrown horizontally from 2.4m above the ground at 35m/s. the wall is 14 m away and 1 m high . at what height the stone will reach? where will be the land?

Answer 1

A stone is thrown horizontally from 2.4m above the ground with speed 35 m/s.

time taken to reach a point which is located 14m away from the stone, t = 14/35 = 0.4s

vertical displacement covered by stone during 0.4 sec , $y=u_yt+\frac{1}{2}a_yt^2$

here, $u_y=0,a_y=-g$

so, y = 0 - 1/2 × g × (0.4)²

if g = 10 m/s²

then, y = -1/2 × 10 × 0.16 = -0.8m

so, position of stone from the ground = 2.4m - 0.8m = 1.6 m

but it is given that, height of wall is just 1m.

so, stone doesn't hit the wall.

now again applying formula, $y=u_yt+\frac{1}{2}a_yt^2$

where, y = -2.4 m , $u_y=0,a_y=-g$

or, -2.4 m = 0 - 1/2 × 10 × t²

or, t² = 0.48 ⇒t = 0.7sec

so, x = ut = 35m/s × 0.7s = 24.5 m

hence, stone hit the ground 24.5 m away from the initial position of it.