A tennis player hits a ball at ground level, giving it an initial velocity of 23

Question

A tennis player hits a ball at ground level, giving it an initial velocity of 23.0 at 60.0 above the horizontal.
What are the horizontal and vertical components of the ball's initial velocity?
How high above the ground does the ball go?
How long does it take the ball to reach its maximum height?
What is the ball's velocity at its highest point?
What is the ball's acceleration at its highest point?
For how long a time is the ball in the air?
When this ball lands on the court, how far is it from the place where it was hit?

Explanation / Answer

Horizontal component of Velocity = vcos(60) = 11.5 m/sec

Vertical component of Velocity = vsin(60) = 19.92 m/sec

Thus v^2 - u^2 = 2as

Thus s = 19.92^2/19.6 = 20.245 m

Velocity at the Heighest point = 0 m/sec

Balls acceleration at heighest point = 9.8 m/sec

 

Let Tbe the time for which ball in air

 

Time taken to attain highest point let be t sec

 

v = u+ gt

 

0 = 19.92 - 9.8*t

 

thus  t = 2.03 sec

 

Same time it takes to come down

 

Thus

T = 2t = 4.06 sec

 

 

Horizontal distance traveled = 4.06*11.5 = 46.7 m

 

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