Question

A ball is thrown toward a cliff of height h with a speed of 25m/s and an angle of 60? above horizontal. It lands on the edge of the cliff 4.0s later.

A.

How high is the cliff?

Express your answer to two significant figures and include the appropriate units.

B.

What was the maximum height of the ball?

Express your answer to two significant figures and include the appropriate units.

C.

What is the ball's impact speed?

Express your answer to two significant figures and include the appropriate units.

Answer 1

a) first find the vertical velocity = 25sin 60 = 21.65m/s

find time to maximum height from v = u+ at
0 = 21.65 - 9.8t
t = 2.20sec

the height reached (answer to b)
v^2 = u^2 + 2as
0 = 21.65^2 - 2 x 9.81 s
s = 23.91m (have to ignore the height of the thrower)

the ball must drop for 4- 2.20 sec = 1.8s
find the height it falls s = 1/2 at^2
s = 0.5 x 9.8 x 1.8^2
s = 15.876m

height of cliff = 21.65 - 15.876 = 5.74m

b) height = 21.65m

c) v = u+at
v = 9.8 x 1.8= 17.64m/s (vertical)