Question

A projectile is shot from the edge of a cliff at an initial speed of 5.0 m/s at angle 30o with the horizontal. After 12 seconds, the projectile hits the ground below at some point P. (a) Determine the height of the cliff above ground level. (b) Determine the distance of the point P from the base of the vertical cliff. (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity

Answer 1

(a)

The initial horizontal velocity of the projectile is

The initial vertical velocity of the projectile is

Let h be the height of the cliff. The initial height of the projectile is y_o=h.

The projectile reaches the ground (y=0) at time t=12 seconds.

The vertical acceleration of the projectile is a_y=-g=-9.8m/s². The vertical acceleration of the projectile is constant, the vertical motion of the projectile is governed by the kinematics equations for uniformly accelerated motion. To find the height of the cliff we use

(b)

The horizontal acceleration of the projectile is zero. The horizontal velocity does not change with time. The distance of point P from the base of the cliff is equal to the horizontal distance traveled by the projectile during the motion.

(c)

The horizontal velocity does not change with time. The horizontal velocity of the projectile just before hitting the ground is the same as the initial horizontal velocity.

The initial vertical velocity of the projectile is . The vertical acceleration is constant, the vertical motion is governed by the kinematics equations for uniformly accelerated motion. To find the vertical velocity of the projectile after time t we use