Projectile Motion Question: Tennis Serve and Net Distance

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a tennis serve. The player serves the ball horizontally and at an angle below the horizontal, with specific parameters regarding the height of the serve and the distance to the net.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations to determine the time of flight and the vertical position of the ball when it reaches the net. There are questions regarding the correct interpretation of acceleration and the setup of the equations.

Discussion Status

Some participants have successfully calculated the time of flight and the vertical position of the ball for part (a) of the problem, while others are exploring the calculations for part (b). There is acknowledgment of confusion regarding the angle of the serve and its implications on the calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on ensuring the correct interpretation of angles and directions in the context of projectile motion.

MFlood7356
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1. During a tennis match, a player serves the ball at 29.4 m/s, with the center of the ball leaving the racquet horizontally 2.41 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.

2. I know that x = x_0 + v_0t + 1/2at^2(this can be applied in the y direction as well), y=(v_0sin(theta))t - (gt^2)/2

3. I've looked at the problem for a good thirty minutes and I've gotten no where. Please help.
 
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Welcome to PF!

Hi MFlood7356! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

Yes, use those equations, and remember that a = 0 for the x-direction …

you know x, so find t, then find y.

What do you get? :smile:
 
Ahh okay I was confused on what the acceleration was because I was having a hard time picturing the problem. I ended up getting 0.408s for time. Now when solving for y I use -9.8 as the acceleration correct?
 
Okay I used y = y_0 + v_0t + 1/2at2 to get 1.59m then 1.59m - 0.900m to get 0.694. Is that right?
 
I put in 0.694m as my answer and that is correct. I attempted to go through and do part b but when I entered the answer it was incorrect. Here's what I did.

Vy= 29.4m/s(sin5) = 2.562 Vx = 29.4m/s(cos5) = 29.288
t = 12/29.288 = 0.410s
y = 2.41m + 2.562(0.410) - 1/2(9.8)(0.410)2 = 2.63673 - .900 = 1.74m
 
MFlood7356 said:
Iy = 2.41m + 2.562(0.410) - 1/2(9.8)(0.410)2 = 2.63673 - .900 = 1.74m

"below the horizontal" :wink:
 
Would that be negative 5 degrees then?
 
Of course. :smile:
 
Okay I got the correct answer thank you!
 

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