What is the angle of projection?

[cryptoguy]

[SOLVED] simple projectile motion..

Homework Statement

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?

Homework Equations

whole bunch for proj motion.

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.

 

[Doc Al]

I know that Dx = 3Dy. Let "t" be the time it takes for proj to reach max height.

When Dy is max, Dx is only half the range.

Dy = Vy*t (vertical)

The vertical motion is accelerated, not constant speed.

3Dy = 2Vx*t (horizontal)

OK.

 

[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 doesn't help me solve Vy/Vx because I get (9.8t^2)/(3Dy). Thank you for any help

 

[Doc Al]

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.

 

[cryptoguy]

aha got it thank you for your help.