Projectile angle and maximum height

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SUMMARY

The discussion centers on determining the projection angle at which the range of a projectile equals its maximum height. Key equations include the maximum height formula, \( V_{fy}^2 = V_{oy}^2 + 2(a)(d) \), and the range formula, \( R = \frac{V_0^2 \sin^2(\theta)}{g} \). Participants emphasize the importance of understanding velocity components and suggest solving for maximum height and equating it to the range. The use of double angle identities is debated, with a focus on simplifying the problem without them.

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ThomasMagnus
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At what projection angle will the range of a projectile equal its maximum height?


I am having a lot of trouble with this one. Is there any way to solve this question without using double angle identities?

I know that this should be the first step:

Max height when Vy=o
Vfy2=Voy2 + 2(a)(dy)

0=(Vo2 + 2(a)(d)

Range= Vo2 sin2(theta)/g


I'm stuck here. Can anyone help me?

Thanks!
 
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What's wrong with double angle identities? They can always be undone. What does sin(2θ) become?

You're on the right track. Solve for the maximum height of the trajectory and set it equal to the range. Be sure to keep track of what's the total velocity and what are components.
 

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