A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor at an angle of 60° above the horizontal, what must the initial speed be (in m/s) if it were to go through the basket?

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t  eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5  eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6  eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

0.866(10.6) = 4.9t² + 1.1

9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

Finally, initial speed can be found by substituting the value of t into eq. 3

Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

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