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

Explanation

A stone is thrown horizontally from 2.4m above the ground with a speed of 35 m/s. time is taken to reach a point which is located 14m away from the stone, t = 14/35 = 0.4s vertical displacement covered by a stone during 0.4 sec, here, 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 the height of the wall is just 1m. so, the stone doesn't hit the wall and the stone is 1.6m above the ground, where the wall stands. now again applying the formula, where, y = -2.4 m , 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, the stone hits the ground 24.5 m away from the initial position of it.