What is the angle of projection?

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The discussion focuses on determining the angle of projection for a projectile whose horizontal range is three times its maximum height. The user initially calculates the ratio of vertical to horizontal velocity components but arrives at an incorrect angle of 34 degrees instead of the correct 53.1 degrees. The mistake is identified in the understanding of vertical motion, which is accelerated rather than uniform. The user receives hints to consider average speed during the ascent to maximum height to resolve the calculations. Ultimately, the correct angle of projection is confirmed to be 53.1 degrees.
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[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.
 
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cryptoguy said:
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.
 
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
 
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.
 
aha got it thank you for your help.
 

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