Question

A basketball player shoots toward a basket 5.6 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 o above the horizontal, what must the initial speed be if it were to go through the basket?

Answer 1

Answer:

9.58 m/s

Explanation:

x = 5.6 m

y = 3 m

θ = 60°

Let u be the initial velocity of projection.

The general equation of projectile path is given by

$y = x tan\theta -\frac{gx^{2}}{2u^{2}Cos^{2}\theta }$

$3 = 5.6\times tan60 -\frac{9.8\times 5.6 \times 5.6}{2\times u^{2}Cos^{2}60}$

$\frac{614.66}{u^{2}}=6.7$

u = 9.58 m/s

thus, the velocity of projection is given by 9.58 m/s.