Projectile Motion of a Tennis Ball crossing over the net

AI Thread Summary
A tennis ball is served at 23.4 m/s from a height of 2.31 m, with the net 12 m away and 0.9 m high. The calculations show that the ball takes approximately 0.513 seconds to reach the net, and its height at that point is 3.60 m, resulting in a distance of 2.7 m above the net. When served at a 5.00° downward angle, the time to reach the net is slightly longer, and the height calculation yields a distance of 1.02 m above the net. The results indicate that in both scenarios, the ball clears the net comfortably. Understanding projectile motion is crucial for analyzing such sports dynamics.
alexadrienne
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Homework Statement



During a tennis match, a player serves the ball at 23.4 m/s, with the center of the ball leaving the racquet horizontally 2.31 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.


Homework Equations



y-y0=vy0cosx+1/2at^2
x-x0=vx0sinx+1/2at^2



The Attempt at a Solution



x-x0=v0xt
12=(1)(23.4)t
t=.513sec

y-y0=voyt+1/2at^2
y=2.31+1/2(-9.8)(.513^2)
y=3.60m

x-x0=v0xt
x-x0=vox(cosx)t
12=23.4cos(5)t
t=.514sec

y-y0=voyt+1/2at^2
y=v0ysin(5)+1/2at^2
y=2.31
 
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y=2.31+1/2(-9.8)(.513^2)
y=3.60m
I get 1.02 for that!
 
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