Question

A basketball player shoots from beyond the 3-point arc. The ball leaves the band with an initial velocity of 8m/s angled 52° from the horizontal. What are the horizontal and vertical velocities of the basketball?

Answer 1

The horizontal velocity (Vx) can be found using the equation: Vx = V0 * cos(theta) where V0 is the initial velocity and theta is the angle from the horizontal.
So, Vx = 8m/s * cos(52°) = 4.74m/s

The vertical velocity (Vy) can be found using the equation: Vy = V0 * sin(theta)
So, Vy = 8m/s * sin(52°) = 6.35m/s

So, the horizontal velocity of the basketball is 4.74m/s and the vertical velocity is 6.35m/s.