During a tennis match, a player serves the ball at 28.2 m/s, with the center of
ID: 1417650 • Letter: D
Question
During a tennis match, a player serves the ball at 28.2 m/s, with the center of the ball leaving the racquet horizontally 2.55 m above the court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number. ( I have tried answering this question, but apparently my answers where wrong. I even checked my work, so I'm super stumped here)
Explanation / Answer
consider the motion of the ball in the horizontal direction
Vox = initial velocity = 28.2 m/s
X = distance travelled to reach above the net = 12 m
t = time taken
using the equation
t = X/Vox = 12/28.2 = 0.43 sec
Consider the motion in Y-direction
Voy = initial velocity = 0 m/s
a = acceleration = - 9.8 m/s2
Yo = initial position = 2.55 m
Yf = final position = ?
using the equation
Yf = Yo + Voy t + (0.5) a t2
Yf = 2.55 + (0) (0.43) + (0.5) (-9.8) (0.43)2
Yf = 1.64 m
distance between ball and top of net = Yf - 0.9 = 1.64 - 0.9 = 0.74 m
b)
consider the motion of the ball in the horizontal direction
Vox = initial velocity = 28.2 Cos5 = 28.1 m/s
X = distance travelled to reach above the net = 12 m
t = time taken
using the equation
t = X/Vox = 12/28.1 = 0.43 sec
Consider the motion in Y-direction
Voy = initial velocity = - 28.2 Sin5 = - 2.46 m/s
a = acceleration = - 9.8 m/s2
Yo = initial position = 2.55 m
Yf = final position = ?
using the equation
Yf = Yo + Voy t + (0.5) a t2
Yf = 2.55 + (- 2.46) (0.43) + (0.5) (-9.8) (0.43)2
Yf = 0.59 m
distance between ball and top of net = Yf - 0.9 = 0.59 - 0.9 = - 0.31 m
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