Projectile motion problem...HELP!

Did you delete the previous post, or did the Mods?

Typed a big-ass explanation then couldn't reply....

I'm so sorry about that! I don't know what happened. I was editting it and then when I went to look back at it, it wasn't there. I appreciate your willing to help. Help me out if you get another opportunity. Thank you so much.

Solve this one the same way you solve the others you posted. Motion in the X direction (distance along the ground) and the Y direction (height) are independent. You use the sine of the angle above horizontal to determine distance in the Y direction and the cosine to get the X direction component.

Vx(time=0) = Cos(37deg)*250 m/s
Vy(time=0) = Sin(37deg)*250 m/s

To get height, solve just the Y portion of motion. The maximum height is when Vx = 0 and the formula for Vx(t) = Vx(0) + at where a is the gravitational constant (and is negative) and t is time. Once you have t, the height is Vy(0)t + 1/2 a t^2 (once again, a is negative).

At this point, you can solve the above equation for height to find the point where height is once again zero, or just remember that it takes the same amount of time to go up as it does to go down. Now you know the total time the ball is in the air.

Finally, multiple the velocity in the X direction by time to get the distance the projectile traveled.

Thanks for your insight! It was a lot of help!