Solving for the Projection Angle: Range and Maximum Height Relationship

[chitownfireman]

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

 

[cristo]

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.

 

[chitownfireman]

Sorry about that mate, thanks for the advice.

 

[saiaspire]

the angle is 76

 

[chitownfireman]

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.

 

[saiaspire]

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...

 

[rocomath]

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.