A basketball player who is 2.00 m tall is standing on the floor 8 m from the basket. If he shoots the ball at a 390 angle with the horizontal, what initial speed must he shoot the ball so that it goes through the hoop without striking the blackboard. The basket height is 3.05 m.

Answer 1

recall that the ball follows the path

y = tanθ x - g/(2 (vcosθ)^2) x^2

plugging in your numbers, then, we need to solve

tan39° * 8 - 4.9/(v cos39°)^2 * 64 = 3.04
6.478 - 519.243/v^2 = 3.05
v = 12.307 m/s

Answer 2

incorrect, i found the correct answer to be 9.78

Answer 3

Ah. I see I forgot to include the initial height of 2 meters.

I assume you discovered that, and made the required adjustment to the equation.