physics - velocity?

Question

A long jump shot is released 2.1 m above the ground, 5.96 m from the base of the basket, which is 3.05 m high. For launch angles of 30° and 60°, find the speed needed to make the basket.

Answer 1

The launch angle is theta. The vertical component of the original launch velocity is
Vov = Vo*sin(theta)
The time to the altitude of the basket:
y = Vov*t + (1/2)*(-g)*t^2
where y = (3.05 - 2.1) m.

Also, that same anount of time applies to the horizontal component of the original launch velocity.
Voh = Vo*cos(theta) = 5.96 m / t

You have 2 equations and 2 unknowns. You can do the math.