Help Me Understand: Homework on Tennis Ball Motion

[whyorwhynot]

I feel like giving up on learning this, really. I don't understand why I can't understand this, yet it seems so simple.

Homework Statement

Suppose that you loft a tennis ball over the net with an initial speed of 15.0 m/s, at an angle of 50 degrees above the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is 2.10 m above its launch point. With what average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)

Homework Equations

The Attempt at a Solution

Vx = (cos 50)(15) = 9.6 m/s
Vy = (sin 50)(15) = 11.5 m/s

Horizontal:
Vx = 15.0 m/s
x = 10.0 m

Vertical:
a = -9.8 m/s²
Vy = 11.5 m/s

I don't know where to go from there.

Thank you if you can help me.

 

[Kurdt]

First you will need to find how far horizontally the tennis ball moves by the time it reaches 2.1 meters above the launch position. Secondly you will need to compare this to the distance the opponent starts out at and the distance he wants to be at (the distance of the ball you just worked out). Then you can work out how long he has to move that distance and the speed he should move.

 

[baba89]

Dear student,

I understand that you are struggling with understanding the concept of tennis ball motion. It is important to remember that physics, like any other subject, requires practice and patience to fully comprehend. Do not get discouraged and keep working on it.

To solve this problem, we can use the equations of motion and the principles of projectile motion. Firstly, we can calculate the time it takes for the ball to reach a height of 2.10 m using the equation: y = y0 + v0y t - 1/2 gt^2, where y0 is the initial height (0 m), v0y is the initial vertical velocity (11.5 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time. Solving for t, we get t = 0.66 s.

Next, we can calculate the horizontal distance traveled by the ball in 0.66 s using the equation: x = x0 + v0x t, where x0 is the initial horizontal position (10.0 m), v0x is the initial horizontal velocity (9.6 m/s), and t is the time. Solving for x, we get x = 16.0 m.

Now, we can use the equation of average velocity, v = (xf - xi)/t, where xf is the final horizontal position (16.0 m), xi is the initial horizontal position (10.0 m), and t is the time (0.66 s). Solving for v, we get v = 9.09 m/s. This is the average speed at which your opponent must move to reach the ball at the desired height.

I hope this explanation helps you understand the problem and the steps involved in solving it. Keep practicing and don't give up. You will eventually get the hang of it. Good luck!