Recent content by DeathEater

well vi*cosθ * (2.477*2) = 53 gives that vi = 10.698/cosθ and could I use the information about the max height? doing that I end up with: 2.477= (vi*sinθ)/g multiply both sides by g and get that 24.3/sinθ = vi setting the two equations equal : 10.698/cosθ = 24.3/sinθ and get...

I reworked it again and got 2.34, my bad. Is that even correct?

I have and I still don't know

I don't know I'm confused as to why you can't just use inverse tan

Well how would I find the launch angle then?

I used the equation for time to fall which is 2.622 = √(2*h/g)

Okay, I tried a different approach. I know that at t1= 2.473 s, the man has a horizontal displacement of 25 m. Since the peak will be reached when he is 26.5 m away, I set up an equality expression and got that t2 (time at peak) is 2.622 s using the formula for time to fall, I found that the...

took cos-1(((125/53)-5)/6) and after recalculating got 116°. Can you help me figure out where I am going wrong?

but the θ came out as -7.33°?

I see my mistake. I am an idiot haha. I got t = 2.34 s (approx)

but I still end up with vi and t, so 2 variables? what do I do about that?

do you mean re-writing vi*cos(θ)*t = 25m & vi*sin(θ)*t -½*g*t2 = 30 m to find cosθ and sinθ and plug that in?

well when I tried it I got θ= 116.12°...so.

yeah I see it, I just don't understand how to get rid of the vi2sin(2θ). You say to replace it with a t, but I don't know what you mean by that. I know the range equation is t(final) * vi*cosθ, but the t's in the other equations aren't the same

yes I see the issue (see my newest comment above). The math just seems a bit "ugly" for 1st year physics so I assumed I was doing something wrong.