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?
Vox = Vo cos 37,
Voy = Vo sin 37,
y = Voy(t) - (1/2)g(t^2)
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.