26.6m/s = v0
36.5º = θ
g = 9.81m/s^2
A) Find Range:
(v0^2)/g *sin2θ = (26.6^2)/9.81 * sin2(36.5)
= 89.99
I double-checked with the other Range equation (2v0*cosθ * (v0sinθ)/g) so I know I'm doing something wrong. Please help! Thank you
edit: A) Correct answer is 69.0
I calculated
Vy = Voy +ay *t
Vy = 0 -981 * (0.286)^2
Vy = -80.24
Then took the total distance 60cm/-80.24 = t = -0.748
But when checked this answer is incorrect.
Vertical position (x component):
y(t) = y0 +v0 * t -1/2*g*t^2
Position yx(t) = 0cm = 40cm + 0 *t -1/2 * 981cm/s/s * t^2
edit: I replaced v0 = 0 and got t = 0.286s, which is incorrect according to the submission.
This didn't work so I thought maybe by finding the angle of the throw would help...
Physics Forums found me actually! I needed some help understanding a problem and in the midst of search it showed up! I started searching for more time solution-related problems and here I am!