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

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 \[B,{{h}_{2}}=ut-\frac{1}{2}g{{t}^{2}}\] \[{{h}_{1}}+{{h}_{2}}=40\,m\] \[\Rightarrow \]\[ut=400,t=400/50=8\,m\] \[\therefore \]\[{{h}_{1}}=320\,m\]and\[{{h}_{2}}=80\,m\]