wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

Two balls are projected upward simultaneously with speeds 40 m/s and 60 m/s. Relative position (x) of second ball w.r.t. first ball at time t=5 s is :
[Neglect air resistance]

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

The correct option is C 100 m
Let the velocity of projections be v1 and v2.
Relative velocity of ball 2 w.r.t 1, urel=v2v1=6040=20 m/s

Since, both the balls are moving under constant acceleration due to gravity. Therefore, relative acceleration of ball 2 w.r.t ball 1 will be zero.
arel=0

Position of ball 2 w.r.t ball 1,
srel=urelt+12arelt2
srel=20×5
srel=100 m

Hence, option (C) is the correct choice.

flag
Suggest Corrections
thumbs-up
24
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon