Question

A handball is hot toward a wall with a velocity of 13.7 m/s in the forward direction. It returns with a velocity of 11.5 m/s in the backward direction. If the time interval during which the ball is accelerated is 0.021s, what is the handballs average acceleration?

Answer 1

1200 m/s2

Step-by-step explanation:

a = (v-u) / t

where a = acceleration

v = final velocity

u = initial velocity

t = time taken

As velocity is a vector it is having a direction. So as the ball bumps out in the opposite direction which it was thrown, the sign of the velocities must be different from each other.

Consider u = -13.7 m/s

then v = +11.5 m/s

t =0.021 s

Applying in the above equation,

a = (11.5-(-13.7))/0.021 = 25.2/0.021 = 1200 m/s2