Question

A ball is thrown toward a cliff of height h with a speed of 27m/s and an angle of 60? above horizontal. It lands on the edge of the cliff 3.9slater.

Part AHow high is the cliff?

Part bWhat was the maximum height of the ball?

PartC What is the ball's impact speed?

Answer 1

Use the below kinematic equation to find the required height.

                 h = usin60^ot + gt²/2

                 h = (27 m/s)(sin60^o)(3.9 s)- (9.8 m/s²)(3.9 s)²/2

                   = 16.66 m

The maximum height of the ball is,

              H = (usin60^o)²/2g

                 = [(27 m/s)(sin60^o)]²/2(9.8 m/s²)

                 = 27.9 m

The impact speed of the ball is,

               v = sqrt[ (27cos60^o)² + (27sin60^o - 9.8x3.9)²]

                  = 20.06 m/s

                  = 20.1 m/s   (approximately)