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 m/s at 60.0 degrees above the horizontal.

a.What are the horizontal and vertical components of the ball's initial velocity?
b. How high above the ground does the ball go?
c. How long does it take the ball to reach its maximum height?
d.What is the ball's velocity at its highest point?
e.What is the ball's acceleration at its highest point?
f. For how long a time is the ball in the air?
g.When this ball lands on the court, how far is it from the place where it was hit?

Explanation / Answer

Initial velocity u = 23 m/s Angle of projection = 60 degrees (a) Horizontal component of initial velocity ux = u cos                                                                  = (23 m/s) cos60                                                                  = 11.5 m/s Verical component of initial velocity uy = u sin                                                             = (23 m/s) sin60                                                             = 19.9 m/s (b) Maximum height H = (uy)^2/2g                              = (19.9 m/s)^2/(2*9.8 m/s^2)                              = 20.24 m (c) Time of ascent t = uy/g                          = (19.9 m/s)/(9.8 m/s^2)                          = 2.03 s (d) The ball's velocity at its highest point = ux = 11.5 m/s (e) The ball's acceleration at its highest point a = g = 9.8 m/s^2 (f) Time of flight T = 2 * time of ascent                           = 2 * t                           = 4.06 s (g) Horizontal distance or Range R = ux * T                                                 = (11.5 m/s)(4.06 s)                                                 = 46.69 m                                         

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