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/*.

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

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

Let uv 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 uy=0.

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

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

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

vx = ux = 15 m/*.

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