Question

Q ) A ball is dropped from a height. of 125m. The acceleration of the ball is 10m/s. Find the time taken by the ball to reach the ground. What is the velocity of the ball after 3s?

(Hint : use 2nd and then 1st equation of motion)​

Answer 1

Given:-

• Height ,h = 125m

• Acceleration ,a = 10m/s²

• initial velocity ,u = 0m/s

• Time taken ,t = = 3s

To Calculate:-

• Time taken by the ball

• Velocity of ball after 3s

Solution:-

By using 1st equation of motion

→ v = u+at

Substitute the value we get

→ v = 0 + ×10×3

→ v = 30m/s

Therefore, the initial velocity of ball after 3s is 30m/s.

Now using 2nd equation of motion

→ s = ut+1/2at²

Substitute the value we get

→ 125 = 0×t + 1/2×10×t²

→ 125 = 5×t²

→ t² = 125/5

→ t² = 25

→ t = √25

→ t = 5s

Therefore, the time taken by the ball to reach the ground is 5 second.