Question

A hot air balloon is rising upward with a constant speed of 20.0 m/s. When the balloon is 20.0 m above the ground, the balloonist accidentally drops a cup over the side. The balloon continue to move up at a constant speed of 20.0 m/s. Ignore the air resistance.

(1) What is the distance between the cup and balloon 1 seconds after it is dropped?

(2) How long does it take for the cup to reach the maximum height?

Answer 1

Given is:-

Initial speed of the balloon or cup is   

Height of the balloon when cup drops

Now,

The distance covered by the cup in 1 second after it is dropped is given by second equation of motion

t2 20

by plugging all the values we get

which gives us

and the distance travelled by balloon during this 1 second is = 20m

therefore the distance between the balloon and the cup after 1 second will be

Part-2

When the cup reach it's maximum height then it's vertical velocity will be zero therefore using first equation of motion we get

by plugging all the values we get

which gives us

Hence,

the cup will take 2.041s to reach its maximum height after it is being dropped.