A) 2
B) 3
C) 4
D) 5
Correct Answer: C
Solution :
Given, \[H=10m\], \[u=5m/s,\text{ }g=10m/{{s}^{2}}\] Speed on reaching ground \[v=\sqrt{{{u}^{2}}+2gh}\] Now, v = u + at Now, \[v=\sqrt{{{u}^{2}}+2gh}=-u+gt\] Time taken to reach highest point is \[t=\frac{u}{g}\], \[\Rightarrow t=\frac{u+\sqrt{{{u}^{2}}+2gH}}{g}=\frac{nu}{g}\text{(from question)}\] \[\Rightarrow 2gH=n\left( n-2 \right){{u}^{2}}\] \[\Rightarrow m\left( n-1 \right)=\frac{2gH}{{{u}^{2}}}=\frac{2\times 10+10}{5\times 5}=8\Rightarrow n=4\]
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