Kinematics

JEE Main level, single correct, medium

66 of 308

Question 66 of 308

From the top of a 6464 metres high tower, a stone is thrown upwards vertically with the velocity of 48m/s.48m/s. The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration g=32m/s2g=32m/{s}^{2}, is:

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Answer: (D) 100100

We know that v2=u2+2gs.{v}^{2}={u}^{2}+2gs.
At greatest height v=0v=0
u=48m/sec⁡u=48m/\sec
g=−32m/sec⁡2g=-32{m/\sec }^{2}
S=48×482.32=36mS=\dfrac{48\times 48}{2.32}=36m
Total distance =64m+36m=100m=64m+36m=100m
=100m=100m

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