You throw a basketball towards a basketball hoop that is 4.92 m away from you, and 0.457 above the release point at 30 degrees above the horizontal (assume g = 10 m/s2)
How fast does the ball have to leave your hand to pass through the hoop on its way down?
So dx = 4.92 m, and dy= 0.457 m
Vx(t) = (Initial velocity x cos30) t
Vy(t) = (Initial velocity x sin30) x gt
The Attempt at a Solution
I know that I just need to find either the x component or y component of velocity as I have θ = 30° which I can use to find the velocity from the release point.
I feel like this question is missing something. I just can't find the time it takes to reach the top (and then it will come back down and go into the hoop) or time it takes to reach the hoop.
Any help would be appreciated.