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?

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15.9,21.8

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Part B

How high above the ground does the ball go?

24.3

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Part C

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

2.5

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Part D

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

15.9

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Part E

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

9.80

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Part F

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

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Part G

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

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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.

vh, vv =

15.9,21.8

  m/s, m/s  

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Correct

Part B

How high above the ground does the ball go?

h =

24.3

  m  

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Correct

Part C

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

t =

2.5

  s  

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Incorrect; Try Again

Part D

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

v =

15.9

  m/s  

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Correct

Part E

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

a =

9.80

  m/s2  

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Correct

Part F

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

t =   s  

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Part G

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

d =   m  

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Explanation / Answer

a) x- component ux = 27 cos(54) = 15.87 m/sec
and y - component uy = 27 sin(54= 21.84 m/sec

b) max height reached H = u^2 sin^2(theta) / 2 g = 27x27 x 0.809 x809 / 19.6 = 24.34 m (nearly)
c) time of ascent ta = u sin(theta) / g = 27 x0.809 / 9.8 = 2.228 sec
e) at the highest point acceleration a = g = 9.8 m/sec^2 (downwards)
f) at t = 0.25 sec acceleration a = g = 9.8 m/sec^2 (c0nstant) but V = u sin(theta) - g t
ie., V = 27x0.809 - 9.8 x0.25 = 19.393 m/sec
g) horizontal distance ie., range R = u^2 sin(2 theta) / g = 27*27 x 1.618/ 9.8 = 120.35 m

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