question_answer

A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity \[25\,\,m{{s}^{-1}}.\] Then the distance from the top of the tower, at which the two balls meet is

A) 68.4 m 

B)        48.4 m

C) 18.4 m 

D)        78.4 m

Correct Answer: D

Solution :

[d] Let the two balls P and Q meet at height x m from the ground after time t s from the start. We have to find distance, \[BC=(100-x)\] For ball P \[S=xm,u=25\,\,m{{s}^{-1}},\] \[a=-g\] From \[S=ut+\frac{1}{2}a{{t}^{2}}\] \[\Rightarrow \]\[x=25t-\frac{1}{2}g{{t}^{2}}\]                           ?(i) Adding eqns. (i) and (ii), we get \[100=25t\] or \[t=4\,s\] From eqn. (i), \[x=21.6\,m\] Hence distance from the top of the tower \[=(100-x)\,m=(100-21.6m)=78.4\,m\]