CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 ms1. The two stones will meet after

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

The correct option is A 4 s
Ordinary method: Using kinematics:

Let the distance from top of the tower where the stones meet be x,

For first stone taking downward as positive we have from the equation of motion

S=ut+12gt2

Here, S=x and u=0

So, for the first stone the point where the two stones meet is at

x=12gt2(1)

Fo the second stone if we take the upward direction to be positive we have

S=ut+12gt2

Here S=100x and u=25 m/s

For the second stone the point where the two stones meet is at

100x=25t12gt2(2)
Adding (1) and (2), we get

25t=100 or t=4 s

Alternative method: Using Relative velocity

Initial velocity of stone 1 which is dropped from the top of a tower, u1=0

Initial velocity of stone 2 which is thrown upward from the ground, u2=+25 m/s
(Take upward direction to be positive)

Hence,
urel=u21=u2u1=250=25 m/s

srel=100 m

arel=gg=0

Using second equation of motion, we get

Srel=urelt+12arelt2

100=25×t+0

t=4 s

Hence, the two stones will meet after 4 seconds.

flag
Suggest Corrections
thumbs-up
0
mid-banner-image
mid-banner-image
similar_icon
Related Videos
thumbnail
lock
Relative Motion
PHYSICS
Watch in App