1

Question

A hot air balloon is rising vertically upwards at a constant velocity of 10 msâˆ’1. When it is at a height of 45 m from the ground, a man bails out from it. After 3 s he opens his parachute and decelerates at a constant rate of 5 msâˆ’2. After how long does the parachutist takes to hit the ground after his exit from the balloon?

Open in App

Solution

The correct option is **B** 5 s

When the parachutist bails out, he has the velocity of the balloon and has an upward velocity of 10 ms−1,i.e.u=+10 ms−1.

Also g=−10 ms−2 (acting downwards).

The displacement of parachutist in t=3 s is given by

S=ut+12gt2 (By second equation of motion)

=10×3+12×(−10)×(3)2

=−15 m

Since the displacement is negative, it is directed downwards.

So the height of the parachutist from the ground when he opened his parachute =45–15=30 m.

The velocity of the parachutist 3 s after he bails out is

u2=u+gt (Using first equation of motion)

=10+(−10)×3=−20 ms−1 (directed downwards)

At t=3 s, the velocity of parachutist is u2=−20 ms−1

Let v2 be the velocity with which the parachutist will reach the ground.

Using third equation of motion, we know

v22−u22=2aS

⇒v22−(−20)2=2(+5)(−30)

⇒v22=400−300=100

⇒v2=−10 m/s (directed downwards)

Let t be the time taken by the parachutist to reach ground after he opens up his parachute Using first equation of motion, we can say

v2=u2+at

⇒−10=−20+(5)t

⇒t=2 s

∴ The total time the parachutist takes (after his exit from the balloon) to hit the ground is =3 s+2 s=5 s

Thus, the correct choice is (b).

When the parachutist bails out, he has the velocity of the balloon and has an upward velocity of 10 ms−1,i.e.u=+10 ms−1.

Also g=−10 ms−2 (acting downwards).

The displacement of parachutist in t=3 s is given by

S=ut+12gt2 (By second equation of motion)

=10×3+12×(−10)×(3)2

=−15 m

Since the displacement is negative, it is directed downwards.

So the height of the parachutist from the ground when he opened his parachute =45–15=30 m.

The velocity of the parachutist 3 s after he bails out is

u2=u+gt (Using first equation of motion)

=10+(−10)×3=−20 ms−1 (directed downwards)

At t=3 s, the velocity of parachutist is u2=−20 ms−1

Let v2 be the velocity with which the parachutist will reach the ground.

Using third equation of motion, we know

v22−u22=2aS

⇒v22−(−20)2=2(+5)(−30)

⇒v22=400−300=100

⇒v2=−10 m/s (directed downwards)

Let t be the time taken by the parachutist to reach ground after he opens up his parachute Using first equation of motion, we can say

v2=u2+at

⇒−10=−20+(5)t

⇒t=2 s

∴ The total time the parachutist takes (after his exit from the balloon) to hit the ground is =3 s+2 s=5 s

Thus, the correct choice is (b).

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program