Kinematics Question 237

Question: A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower [CPMT 2003]

Options:

A) 100 meters

B) 320 meters

C) 80 meters

D) 240 meters

Show Answer

Answer:

Correct Answer: C

Solution:

Let both balls meet at point P after time t. The distance travelled by ball A, $ h _{1}=\frac{1}{2}gt^{2} $ The distance travelled by ball B, $ h _{2}=ut-\frac{1}{2}gt^{2} $

$ h _{1}+h _{2}=400\ m $

therefore $ ut=400,\ t=400/50=8\ sec $
$ \therefore $ $ h _{1}=320\ m\ And\ \ h _{2}=80\ m $



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