What Angle of Projection Makes Maximum Height Equal to Range?

[kuakkgom]

This problem has been giving me headaches for the past two weeks. Here it is, as well as how far I have gotten on it:

A projectile is fired with an initial velocity of 53m/s. Find the angle of projection such that the maximum height is equal to it's range.

So that basically boils down to:

H = R
Vo = 53m/s
ay = -9.81m/s^2
t = unknown
x (angle) = find

According to the nifty sheet the teacher tossed out the applicable formulas would be:

R=Vo^2*sin(2x)/g (since Ho and Hf will be the same)
H=Voy^2/2g.

Making H=R gets the formula Voy^2=2*Vo^2*sin(2x) ...which doesn't help me since x is what needs to be found, and without x one can't break Vo into component form.

Do I have the right idea or have I flubbed it already?

Thanks,

-K

 

[Leong]

Voy=vo*sinx

 

[quicknote]

atie

Dear Katie,

It seems like you have the right idea in terms of using the equations for range and maximum height. However, there may be a few things you could adjust to help you solve the problem more easily.

Firstly, instead of using the formula for maximum height (H=Voy^2/2g), it might be more helpful to use the formula for time of flight (t). This is because the time of flight will be the same for both the maximum height and range, since the projectile will reach its maximum height at the midpoint of its trajectory. So we can set H=0 in the formula for time of flight and solve for t.

Once you have t, you can use it to solve for the angle of projection (x) using the formula for range (R=Vo*t*sin(x)). This will give you a more direct way to find the angle without having to rearrange equations.

I hope this helps and good luck with your problem! Remember to always check your units and make sure they are consistent throughout your calculations.

Best,