Question

A ball is projected at an angle of 53 degrees and a speed of 30m/s from the top of a cliff on Venus. The ball lands onto the flat, level ground below the cliff and 100m from the base of the cliff. how tall is the cliff? gravity on venus is equal to 8.87m/s^2

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration on Venus = g = -8.87 m/s²

Initial velocity of the ball = V = 30 m/s

Angle at which the ball is thrown = = 53^o

Initial horizontal velocity of the ball = V_x = VCos = (30)Cos(53) = 18.05 m/s

Initial vertical velocity of the ball = V_y = VSin = (30)Sin(53) = 23.96 m/s

Horizontal distance the ball covers before reaching the ground = R = 100 m

Time taken by the ball to reach the ground = T

There is no horizontal force acting on the ball therefore the horizontal velocity of the ball remains constant.

R = V_xT

100 = (18.05)T

T = 5.54 sec

Height of the cliff = H

When the ball reaches the ground the vertical displacement of the ball is in the downwards direction therefore it is negative.

-H = V_yT + gT²/2

-H = (23.96)(5.54) + (-8.87)(5.54)²/2

H = 3.38 m

Height of the cliff = 3.38 m