Question

Q5 solve the following (2)
A body is released from the top of A
building of height 19.6m
Find the velocity with which the body
hits the ground.​

Answer 1

Answer:

19.6 m/s

Explanation:

Given:

• Initial velocity of the object = u = 0 m/s
• Distance = s = 19.6 metres
• Acceleration = a = 9.8 m/s²

To find:

• Final velocity of the object with which it hits the ground = v

Using the third equation of motion : v²-u²=2as

Substituting the values:

v²-0²=2×9.8×19.6

v²=384.16

v=√384.16

v=19.6 m/s

The final velocity of the body is equal to 19.6 m/s

More:

• First equation of motion = v = u+at
• Second equation of motion = s=ut+1/2 at²