Question

A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the acceleration for the dropped ball.

Answer 1

Explanation:

The given data is as follows.

        Initial velocity; u = 0,        Final velocity; v = 19.6 m/s

        time; t = 2 seconds

As the relation between initial velocity, final velocity and acceleration is as follows.

                         v = u + at

Hence, putting the given values into the above formula as follows.

                       v = u + at

           19.6 m/s = 0 + $a \times 2 sec$

                   a = 9.8 $m/s^{2}$              

Thus, we can conclude that acceleration of the dropped ball is 9.8 $m/s^{2}$.

Answer 2

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²