wiz-icon
MyQuestionIcon
MyQuestionIcon
11
You visited us 11 times! Enjoying our articles? Unlock Full Access!
Question

Two balls are dropped from different heights at different instants. Second ball is dropped 2 s after the first ball. If both the balls reach the ground simultaneously, after 5 s of dropping the first ball, then the difference between the initial heights of the two balls will be
(Take g=9.8 m/s2)

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

The correct option is B 78.4 m

Since, both ball reach ground after 5 sec dropping 1st ball, so the time taken by the 1st ball to reach the ground is t1=5 s.

From diagram, height travelled by 1st ball in 5 s is h1.

So, using equation of motion for 1st ball, we get

h1=12gt21

h1=12g(5)2=252g

Since the 2nd ball is dropped 2 s after the 1st ball, so the time taken by it to reach the ground is (t2=52=3 s).

Similarly, height covered by 2nd ball in 3 s is h2

h2=12g(3)2=92g

So, the difference in the heights,

h1h2=252g92g

h1h2=8g

h1h2=8×9.8=78.4 m

Hence, option (b) is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon