Question

25.A balloon rising vertically up with a velocity of 10m/s releases a sandbag at an
instant when the balloon is 20m above the ground.
(a) How many seconds after its release will the bag strike the ground.
(b) Find the distance covered by the bag. (g=10m/s)​

Answer 1

Explanation:

Given,

u=10m/s in upward direction.

g=−10m/s

2

in downward direction

t=3sec

H=45m

So by second equation of motion:

s=ut+

2

1

at

2

=10×3+

2

1

(−10)×3×3

=30−45

=−15m

Here negative sign indicates that it is directed downwards.

So height from the ground when he opened parachute is =45−15=30m