JEE Main 2015PhysicsMotion In One DimensionEasyMCQ

JEE Main 2015 — Motion In One Dimension Question with Solution

JEE Main 2015 (11 Apr Online)

Question

From the top of a 64 metres high tower, a stone is thrown upwards vertically with the velocity of 48 m/s. The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration g=32 m/s2, is:

Choose an option

Show full solutionCorrect option: D
Correct answer
D100

Step-by-step explanation

We know that v2=u2+2gs.

At greatest height v=0

u=48 m/sec

g=-32 m/sec2

S=48×482.32=36 m

Total distance =64 m+36 m=100 m

=100 m

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About this question

This is a previous-year question from JEE Main 2015, covering the Motion In One Dimension chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.