# Simple projectile motion

1. Dec 22, 2007

### cryptoguy

[SOLVED] simple projectile motion..

1. The problem statement, all variables and given/known data
A projectile is fired in such a way that its horizontal range is equal to 3 times its maximum height. What is the angle of projection?

2. Relevant equations
whole bunch for proj motion.

3. The attempt at a solution
I know that Dx = 3Dy. Let "t" be the time it takes for proj to reach max height.
Dy = Vy*t (vertical)
3Dy = 2Vx*t (horizontal)

Vy/Vx = (2t*dy)/(3t*dy) = 2/3

inverse tan (2/3) = 34 degrees, but the answer is 53.1 degrees. Any ideas what's wrong?
Thanks.

2. Dec 22, 2007

### Staff: Mentor

When Dy is max, Dx is only half the range.
The vertical motion is accelerated, not constant speed.
OK.

3. Dec 22, 2007

### cryptoguy

Ok lol what a stupid mistake about the Dy. But now I'm not sure what to do... I know Vf = Vi + at. So Vy = 9.8t, and it still doesnt help me solve Vy/Vx cuz I get (9.8t^2)/(3Dy). Thank you for any help

4. Dec 23, 2007

### Staff: Mentor

Hint: If Vy is the intial speed (in the vertical direction), find the average speed during the rise to maximum height. Dy will equal average speed x time.

5. Dec 23, 2007

### cryptoguy

aha got it thank you for your help.
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