A tennis player hits a ball at ground level, giving it an initial velocity of 21

ID: 1592919 • Letter: A

Question

A tennis player hits a ball at ground level, giving it an initial velocity of 21.0 m/s at 60.0 above the horizontal.

PART A

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

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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?

here,

velocity of ball, Vo = 21 m/s
Angle, A = 60

Part A:
Vx = Vo * CosA = 21*Cos60 = 10.5 m/s
Vy = Vo * SinA = 21 *Sin60 = 18.187 m/s

Part B:
height in projectile motion is given as ,
h = (Vy)^2/2g
h = (18.187)^2/(2*9.8)
h = 16.876 m

Part C:
Time takne to reach max height,
T = 2*Vy/g
T = 2*18.187/9.8
T = m/s

Part D:
at heighest point balll will have only horizontal component of velocity which is equla to initial horizontal velocity i.e

Vxh = Vx
Vyh = 0

Part E :
at heighest point ball will only have acceleration equal to acceleration due to gravity.
a = 9.81 m/s^2

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