Projectile Motion (Ball thrown against a Wall)

[TheSnacks]

1. The problem statement, all variables and given/known data

You throw a ball toward a wall at speed 24.0 m/s and at angle θ0 = 37.0° above the horizontal. The wall is distance d = 20.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?


2. Relevant equations

Vox = Vo cos 37,

Voy = Vo sin 37,

y = Voy(t) - (1/2)g(t^2)

3. The attempt at a solution

I got the answer for (b) because I just said Vx= 24cos(37), which came out to be 19.1673. The homework is on the computer, so it let me know I was right. I tried to do the same thing for Vy, but using sine and it says I am incorrect.

to get the height as it hits the wall, I figure I could use the equation given to solve for y and get the answer. But I need to figure out Voy first. And I figure that time could just be .833333 seconds because 20m/24(m/s) should give me time, right?

Any help would be great. I must just be looking at the problem wrong.

[Doc Al]

I got the answer for (b) because I just said Vx= 24cos(37), which came out to be 19.1673. The homework is on the computer, so it let me know I was right. I tried to do the same thing for Vy, but using sine and it says I am incorrect.
Vy is a function of time; it's not constant like Vx. (You can find Voy that way though.) How does Vy depend on time?

to get the height as it hits the wall, I figure I could use the equation given to solve for y and get the answer. But I need to figure out Voy first.
Yes, so figure out Voy.
And I figure that time could just be .833333 seconds because 20m/24(m/s) should give me time, right?
Not exactly. What's the horizontal velocity?

[TheSnacks]

Well if I try to find Voy out the same way, I get 14.4436. But that isn't correct. Vy depends on time because, as time passes, Vy gets smaller until it reaches 0 and starts to fall again.


Well, the horizontal velocity would be 19.16 then. Right? So that would make the time 1.04 seconds.

[librab103]

Well if I try to find Voy out the same way, I get 14.4436. But that isn't correct. Vy depends on time because, as time passes, Vy gets smaller until it reaches 0 and starts to fall again.


Your vyo of 14.4436 m/s is correct. So use that plus your t= 1.04 to find vy and y.

I am wondering why you are making the assumption that the ball hits the wall on the way down? If you find vy at time t, you will see that it is going much slower then when it was first release but it is still a positive number thus the ball hits the wall as it is still going up in its trajectory.