Question

A basketball is shot from an initial height of 2.40 m (Figure 1)with an initial speed v0 = 13 m/s directed at an angle θ0 = 45 ∘above the horizontal.

A)How far from the basket was the player if he made a basket?

B) At what angle to the horizontal did the ball enter the basket?

Answer 1

(a)

Apply kinematic equation of motion for y - direction,

y = yo + vo*t - (1/2)gt^2

3.05 = 2.4 + 13*sin 45*t - (1/2)*9.8*t^2

4.9*t^2 - 9.19t + 0.65 = 0

By solving above quadratic equation,

t = 1.802 s

x = vx * t

x = 13*cos 45 * 1.802

x = 16.56 m

(b)

vx = 13*cos 45 = 9.192 m/s

vy = vo*sin45 - gt = 13*sin45 - 9.8*1.802

vy = -8.47 m/s

angle $\theta$ = tan^-1 (vy / vx)

$\theta$ = tan^-1 (-8.47 / 9.192)

$\theta$ = -42.66 deg

$\theta$ = 42.66 deg below the horizontal