Question

A body is released from the top of the of height 100 m . After 2 sec it is stopped instantaneously released . What will be it's height after 3 seconds ? [ Take g — 10m/s ^2

Answer 1

Given:-

• Initial velocity of body (u) =0m/s

• Height (h) = 100m

• Time taken (t) = 2 s

• Acceleration due to gravity (g) = 10m/s²

To Find:-

• Height of body after 3 second (h)

Solution:-

As the Body released from the top of the tower .Then the initial velocity of body is 0m/s

By using 2nd equation of motion

→ s = ut +1/2at²

Substitute the value , we get

→ s = 0×3+ 1/2×10×3²

→ s = 0+1/2×10×9

→ s = 1/2×90

→ s = 45m

∴ The height of body after 3 second is 100m - 45m = 55m