question_answer

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]

A)             100 meters      

B)             320 meters

C)             80 meters

D)                      240 meters

Correct Answer: C

Solution :

                Let both balls meet at point P after time t.             The distance travelled by ball A, \[{{h}_{1}}=\frac{1}{2}g{{t}^{2}}\]                The distance travelled by ball B, \[{{h}_{2}}=ut-\frac{1}{2}g{{t}^{2}}\]             \[{{h}_{1}}+{{h}_{2}}=400\ m\] Þ\[ut=400,\ t=400/50=8\ sec\]             \[\therefore \] \[{{h}_{1}}=320\ m\ \ \text{and}\ \ {{h}_{2}}=80\ m\]