A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle

Question

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60 degree above horizontal. It lands on the edge of the cliff 3.3 s later. How high is the cliff? Express your answer to two significant figures and include the appropriate units. What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. What is the ball's impact speed? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

v = 33sin60 = 28.58 m/s
vertical position in time
h = h + vt + ½gt²
h = 0 + 28.58(3.3) + ½(-9.8)3.3²
h = 40.953m

b) What was the maximum height of the ball?

v² = v² + 2as
at max height, vertical velocity = 0
0² = 28.58² + 2(-9.8)s
s = 41.674 m

c) What is the ball s impact speed?
v² = v² + 2as
v² = 33² + 2(-9.8)(40.953)
v =16.92 m/s

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