Ahandball 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?

1200 m/s2

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

no. dot product   =0 when vecs at right angles

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