Yet another Projectile motion question

[Rectophobia]

My physics teacher gave me another challenge question and I figured that since you guys were helpful last time, I might as well ask for input before I hand this one in too. Since my handwriting is pretty illegible, I'll repeat the question. Find the angle, theta, such that the maximum height attained is half of the overall horizontal distance traveled. Assume that y0 = 0 and wind resistance is negligible.


PS. (If my answer is correct). Since I got the answer tan(theta) = 2, I was wondering if there is a faster way to arrive at my answer since it seems like the kind of answer you get from 2 lines of working as opposed to a page.

[availn]

My physics teacher gave me another challenge question and I figured that since you guys were helpful last time, I might as well ask for input before I hand this one in too. Since my handwriting is pretty illegible, I'll repeat the question. Find the angle, theta, such that the maximum height attained is half of the overall horizontal distance traveled. Assume that y0 = 0 and wind resistance is negligible.


PS. (If my answer is correct). Since I got the answer tan(theta) = 2, I was wondering if there is a faster way to arrive at my answer since it seems like the kind of answer you get from 2 lines of working as opposed to a page.

lol, there wasn't anything for us to be helpful with last time, as you were correct then, as you are now. The only way you could really do this faster is to have pre-derived formulas on your cheat sheet for projectile motion questions. For example:

Horizontal distance traveled = u²sin(2θ)
                                                   g

Vertical distance traveled = u²sin²(θ)
                                              2g

From there, you can equate half of the first equation with the second, getting:
sin(2θ) = sin²(θ)
tan(θ) = 2

[Rectophobia]

lol, there wasn't anything for us to be helpful with last time, as you were correct then, as you are now. The only way you could really do this faster is to have pre-derived formulas on your cheat sheet for projectile motion questions. For example:

Horizontal distance traveled = u²sin(2θ)
                                                   g

Vertical distance traveled = u²sin²(θ)
                                              2g

From there, you can equate half of the first equation with the second, getting:
sin(2θ) = sin²(θ)
tan(θ) = 2

Thanks again availn, you're a life saver. You might not have had to help last time with any of the calculations, but the fact that someone who scored so well in physics told me that I did it right gives me a sense of confidence in my ability. So thanks again and I hope you do well this year

[availn]

Thanks again availn, you're a life saver. You might not have had to help last time with any of the calculations, but the fact that someone who scored so well in physics told me that I did it right gives me a sense of confidence in my ability. So thanks again and I hope you do well this year

Best of luck to you too! You should have a ton of confidence in your ability, these questions go beyond what would appear in the exam, and you've gotten through them easily