Question

A body is released from the top of a tower of height H m. After 2sec it is stopped and then instantaneously released. What will be the its height after next 2sec ( in meters)? ​

Answer 1

Given:-

• Initial velocity (u) = 0m/s

• Time taken (t) = 2 s

• Time taken (t') = 2 s

To Find:-

• Height of the body (H).

Solution:-

As we know that the acceleration due to gravity is 9.8m/s²

Height of body after 2sec

By using 2nd equation of motion

→ H = ut +1/2at²

Substitute the value we get

→ H = 0×2+1/2×9.8×2²

→ H = 0+1/2×9.8 ×4

→ H = 9.8×2

→ H = 19.6 m.

Similar

Height attained by body after 2second

→ H = 9.8×2

→ H = 19.6m

∴ Total height attained by the body is 19.6+19.6= 39.2metre.