Projectile Motion Question: Tennis Serve and Net Distance

In summary, 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, the distance between the center of the ball and the top of the net is 0.694 m. When the ball is served at a 5.00° angle below the horizontal, the distance between the center of the ball and the top of the net is 1.74 m.
  • #1
MFlood7356
39
0
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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  • #2
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:
 
  • #3
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?
 
  • #4
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?
 
  • #5
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
 
  • #6
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:
 
  • #7
Would that be negative 5 degrees then?
 
  • #8
Of course. :smile:
 
  • #9
Okay I got the correct answer thank you!
 

FAQ: Projectile Motion Question: Tennis Serve and Net Distance

1. What is projectile motion?

Projectile motion is the motion of an object through the air or vacuum under the influence of gravity alone. It is a type of motion in which the object is only acted upon by the force of gravity, and no other forces such as air resistance are present.

2. What are the key factors that affect projectile motion?

The key factors that affect projectile motion are the initial velocity, the angle of launch, and the force of gravity. These factors determine the trajectory and range of the projectile.

3. How does changing the angle of launch affect the trajectory of a projectile?

Changing the angle of launch affects the trajectory of a projectile by altering the vertical and horizontal components of its velocity. A higher angle will result in a higher peak height and a shorter range, while a lower angle will result in a longer range and a lower peak height.

4. What is the formula for calculating the range of a projectile?

The formula for calculating the range of a projectile is Range = (initial velocity)^2 x sin(2 x launch angle) / gravity. Alternatively, it can also be calculated as Range = (initial velocity)^2 x sin(launch angle) x cos(launch angle) / gravity.

5. Can air resistance affect the trajectory of a projectile?

Yes, air resistance can affect the trajectory of a projectile. In real-world situations, air resistance will act on a projectile and cause it to deviate from its expected path. However, in idealized situations where air resistance is negligible, the trajectory will follow the predicted path based on the initial conditions and the force of gravity.

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