wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A man in a hot air balloon, throws a stone downwards with a speed of 5 m/s with respect to the balloon. If the balloon is moving upwards with a constant acceleration of 5 m/s2, then the velocity of the stone relative to the man after 2 second is
(Take g=10 m/s2)


A
30 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
15 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
10 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
35 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 35 m/s
Let u be the relative velocity of stone with respect to man and a be the relative acceleration of same.

u=5 m/s and
a=5+g=5+10=15 m/s2

Now using first equation of motion

v=u+at

v=5+(15×2)=35 m/s

relative velocity after t=2 second is 35 m/s.

Why this question?Acceleration of projectile w.r.t balloon = Acceleration of projectile w.r.t ground- Acceleration of balloon w.r.t ground.

flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon