Question

During warm ups, a tennis player hits a ball which is 2.0 m above the ground. The ball leaves his racquet with a speed of 16.0 m/s at an angle 4.8 ∘above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Assume that the path of the ball, as observed from a bird's eye view, is parallel to the long lines of the tennis court.

A) At what time, after being struck by the racket, does the ball either pass over the net or else hit the net? (ANSWER IN S)

B) By how much does the ball clear the net? If the ball does not clear the net, but instead hits the net, enter this height as a negative number. (ANSWER IN M)

C) At what time, after being struck by the racket, does the ball reach its maximum altitude? (ANSWER IN S)

D) What is that maximum altitude? (ANSWER IN M)

E) At what time, after being struck by the racket, does the ball finally hit the ground? Note: if the ball would have hit the net, assume that the net has a hole in it at that location so that the ball would pass through the net with no interruption in its flight path. (ANSWER IN S)

F) At what horizontal distance from the net does the ball finally hit the ground? (ANSWER IN M)

Answer 1

(A)

ball have to travel 7 m horizontally .

t = distance / horizontal velocity

t   = 7/16*cos4.8 = 0.43 sec

(B)

in vertical

h = 16*sin4.8 * 0.43 - 9.8 * (0.43)^2 /2
h = 0.575 -0.906= -0.330 m

net is below 2-1 = 1m .

clear height = 1 + (-0.330) = 0.67 m

(C)

max. height when vertical component of velocity   = 0

v_y = u_y +a*t

0 = 16*sin4.8 - 9.8*t
t= 0.136 sec

(D)

h = u_y*t +a*t^2 /2

h = 16*sin4.8 *0.136 - 9.8*(0.136)^2 /2

h = 0.182 - 0.090

h = 0.092 m form racquet

fromground   = 2+ 0.092 = 2.092 m

(E)

to hit ground ball have travel 2 m in vertical downwrads .

h = u*t + at^2 /2

-2 = 16*sin4.8*t - 9.8*t^2 /2

4.9*t^2 - 1.33*t -2 = 0

t = 0.78 sec

(F)

d = u_x*t

d = 16*cos4.8 * 0.78 = 12.43 m