Question

38. A balloon is rising vertically up with a velocity of 29 ms.
A stone is dropped from it and it reaches the ground in
10 seconds. The height of the balloon when the stone was
dropped from it is (g = 9.8 ms (WBJEE 2014]​

Answer 1

Question:-

A balloon is rising vertically up with a velocity of 29 ms.A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it is (g = 9.8 ms).

Formulas:-

• v = u +at
• v² - u² = 2as
• s = ut + 1/2 at²

Solution:-

For a stone which is thrown downwards from a balloon rising upwards, the equation of motion is

s = −ut + 1/2gt²

= -29 × 10 + 1/2 × 9.8 × 10 × 10

=−290 + 490

=200m.

Hence the height of the balloon when the stone was 200m.

Answer 2

Explanation:

Since the Balloon is moving upward the initial velocity of the stone is -29m/s (downward direction is taken to be positive)

acceleration due to gravity= +9.8 m/s² (g is towards downtown direction)

therefore u= -29 m/s

g=9.8 m/s²

t=10 secs

according to the eqns

s=ut+1/2gt²

=( -29×10)+(1/2×9.8×10²)

= -290+490

= 200 m

hence the height of the Balloon when the Stone is dropped is 200 m.