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

acceleration, a = 9.8 m/s²

Step-by-step 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

Answer 2

Step-by-step 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 * 2 sec`

a = 9.8
`m/s^(2)`

Thus, we can conclude that acceleration of the dropped ball is 9.8
`m/s^(2)`.