Projectile Motion of a Tennis Ball crossing over the net

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SUMMARY

The discussion focuses on calculating the projectile motion of a tennis ball served at 23.4 m/s from a height of 2.31 m, with the net positioned 12.0 m away and 0.900 m high. The calculations reveal that when served horizontally, the ball clears the net by 1.02 m. Additionally, when served at an angle of 5.00° below the horizontal, the ball's trajectory results in it also clearing the net, although the exact distance is not specified in the provided calculations. The equations used include kinematic equations for horizontal and vertical motion.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic trigonometry for angle calculations
  • Knowledge of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn about the effects of launch angles on projectile trajectories
  • Explore simulations of projectile motion using tools like PhET Interactive Simulations
  • Investigate real-world applications of projectile motion in sports physics
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Students studying physics, particularly those focusing on mechanics, as well as coaches and athletes interested in understanding the dynamics of tennis serves.

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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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