Question

A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms^–1. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (g = 9.8 ms^–2)

Answer 1

Motion of a stone may be considered as the superposition of the two independent motions. 

Given: Horizontal motion with constant velocity, u = 15 m/s. 

Vertical motion with constant acceleration, a = g = 9.8 m/s2. 

Let h be the height of the cliff above the ground. 

Let u_v be the vertical (downward) component of the velocity of prjection of the stone. 

If the stone hits the ground after t seconds of projection, then

Since the stone is thrown horizontally, the vertical component of velocity u_y =0.

This gives the time t for stone to reach the ground,

Let v_y be the vertical (downward) component of velocity of the stone when it hits the ground, then

The horizontal component of velocity v_x with which the stone hits the ground it remains constant because there is no acceleration in the horizontal direction. 

v_x = u_x = 15 m/s.

Thus, the final speed with which the stone hits the ground is 99.14 m/s.