Question

A balloon rising vertically with a velocity of 5 m/
sec drops a body from a height of 20 m above
the ground. The velocity of the body 1 sec after
release is:
1)4.8 m/sec up 2)4.8 m/sec down
3)9.8 m/sec up 4)9.8 m/sec down​

Answer 1

Answer:

The correct answer is (4) 4.8 m/sec down

Explanation:

Let h be the initial height of the body while it was dropped, u be its initial velocity and v be its velocity after time t, and a be the acceleration of the body.

Given:

h = 20 m

u = 5 m/sec

t = 1 sec

and also we have acceleration due to gravity, g = 9.8 m/s²

To find:

v, that is, the velocity of the body after 1 second of being dropped.

Solution:

Let us take the positive Y-axis as the positive direction by convention.

Thus, the acceleration of the body is a = -g, as gravity acts along the negative Y-axis.

⇒ a = -9.8 m/s²

From the kinematical equation v = u + at, we get

⇒ v = 5 - 9.8(1)

⇒ v = -4.8 m/s

Thus the velocity of the body is 4.8 m/s in the downward direction.