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 \[25\text{ }m{{s}^{-1}}\]. 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 :

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, \[BC=\left( 100-x \right)\] For ball P \[\text{S=x m, u=25 m}{{\text{s}}^{-1}},\text{a}=-\text{g}\] From \[S=ut+\frac{1}{2}a{{t}^{2}}\] \[x=25t-\frac{1}{2}a{{t}^{2}}\]                                 .......... (i) For ball Q \[\text{S= }\left( 100-\text{x} \right)\text{ m, u = 0, a =g}\] \[\therefore 100-x=0+\frac{1}{2}g{{t}^{2}}\]                .......... (ii) Adding eqn. (i) and (ii), we get \[\text{100=25t or t=4s}\] From eqn. (i), \[\text{x=25}\times 4-\frac{1}{2}\times 9.8\times {{\left( 4 \right)}^{2}}=21.6\text{ m }\] Hence distance from the top of the tower \[\text{= (100-x)m=(100-21}\text{.6m)=78}\text{.4m}\]