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...
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...
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