Kinematics Question 237
A man drops a ball downward from the roof of a tower of height 400 meters. At the same time another ball is thrown upward 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\ \text{And}\ h _{2}=80\ m $
BETA
