Solving for the Projection Angle: Range and Maximum Height Relationship

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Homework Help Overview

The problem involves determining the projection angle at which the range of a projectile equals its maximum height. This falls under the subject area of projectile motion in physics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to derive equations for range and maximum height, suggesting a breakdown into vertical and horizontal components. There is also mention of the relationship between sine and cosine functions in the context of projectile motion.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, emphasizing the importance of showing work. There is an acknowledgment of the need for further exploration of the relationships between the variables involved, but no consensus has been reached on the solution.

Contextual Notes

Participants note that solutions are not provided on the forum, and there is a reminder to adhere to homework posting guidelines. There is also a mention of a specific angle being suggested, but this is framed within the context of ongoing exploration rather than a definitive answer.

chitownfireman
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My teacher gave me this problem today and I have tried everything I know but I still haven't found the right answer. If anyone knows how to solve it, please share. Thanks

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


Hint: 2 sin θ cos θ = sin 2 θ
 
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Presumably you've either been given a formula for the range of a projectile, or know the kinematic equations and can derive such an equation. Do the usual thing, split up into vertical and horizontal components, and solve for each separately.

You need to show some work before you get any real help, and should note that solutions are not provided on these forums at all. In future, also, please post in the homework forums. Oh, and welcome to PF.
 
Sorry about that mate, thanks for the advice.
 
the angle is 76
 
Thanks mate, but I never wanted the exact answer, I just wanted a push in the right direction. You may be right, but I am currently trying to figure it out, but thanks for the input.
 
Let A be the angle and u be the velocity of projection and g ( accel due to gravity)
Range (R) = u^2sin A^2/2g

Max Height(H)=u^2sin2 A/g

sin 2 A = 2 sin A cos A

equate R and H ( since they should be equal)..

u will get sin A/cos A = 4

but sin A/cos A = tan A

therfore A should be 76 degrees...
 
saiaspire said:
Let A be the angle and u be the velocity of projection and g ( accel due to gravity)
Range (R) = u^2sin A^2/2g

Max Height(H)=u^2sin2 A/g

sin 2 A = 2 sin A cos A

equate R and H ( since they should be equal)..

u will get sin A/cos A = 4

but sin A/cos A = tan A

therfore A should be 76 degrees...
you're not suppose to give out full solutions or the answer.
 

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