Question

When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the average force was 72.9 N, how much time was it in contact with the floor?
(Unit = s)
Remember: up is +, down is -

Answer 1

Answer:

t = 0.0689 s

Explanation:

Given that,

Mass of a basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s (downward or negative)

Final velocity, v = 3.85 m/s (up of positive)

Average force, F = 72.9 N

We need to find the time it was in contact with the floor. The force is given by :

$F=ma\\\\F=m\dfrac{v-u}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-(-4.23))}{72.9}\\\\t=0.0689\ s$

So, the time of contact is 0.0689 s.