Question

when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how much time was it in contact with the floor?

Answer 1

Answer:

Time, t = 3.2 ms

Explanation:

It is given that,

Mass of basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s

Final velocity, v = 3.85 m/s

Average force acting on the ball, F = 72.9 N

We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms$

So, the ball is in contact with the floor for 3.2 ms.