A tennis player hits a ball at ground level, giving it an initial velocity of 27

Question

A tennis player hits a ball at ground level, giving it an initial velocity of 27.0m/s at 54.0? above the horizontal.

Part A

What are the horizontal vh and vertical vv components of the ball's initial velocity?

Enter your answers separated by a comma.

Part B

How high above the ground does the ball go?

Part C

How long does it take the ball to reach its maximum height?

Part D

What is the ball's velocity at its highest point?

Part E

What is the ball's acceleration at its highest point?

Part F

For how long a time is the ball in the air?

Part G

When this ball lands on the court, how far is it from the place where it was hit?

Explanation / Answer

PART-A

a) vh = 27 cos 54 =15.87m/sec

Vv = 27 sin 54=21.843m/sec

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PART-B

b) max height = Vv^2/2g =21.843^2/2*9.8

Max height = 24.343m
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PART-C

c) t = Vv/g =21.843/9.8

t=2.22sec

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d) at max height, vertical velocity is zero; the only component of velocity at max height is horizontal which is unchanged from 27 cos 54

v=15.84m/sec

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e) while in the air, the accel is always = g = -9.8m/s/s

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f) twice the time to reach max ht

t=2.22*2 =4.44sec

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g) horizontal range = time of flight x horizontal velocity

vh = 27 cos 54 =15.87m/sec

range= 15.87*4.44

=70.46m

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