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 \[=(100-x)\,m=(100-21.6m)=78.4\,m\]You need to login to perform this action.
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