CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance).

Open in App
Solution

Height of the cliff, h = 240 m
Velocity of Stone 1, y1= 10 m/s
Velocity of Stone 2, y2 = 40 m/s
To calculate the time when stone 1 strikes the ground, use Equation of Motion,
h=ut+12gt2
240=10×t12×10×t2
5t210t240=0
t22t48=0
t = 8 sec and -6 sec
To calculate the time when stone 2 strikes the ground, use Equation of Motion,
h=ut+12gt2
240=40×t12×10×t2
5t240t240=0
t=12 sec
The relative velocity of the two stone =y2y1=4010=30 m/s
Therefore, graph of y2y1 against time is a straight line.
Hence, when the stone strikes the ground, the relative velocity will increase up to 12 seconds.



flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Speed and Velocity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon