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Question Number 30331 by Tinkutara last updated on 20/Feb/18

Answered by rahul 19 last updated on 20/Feb/18

$$\left(3\right)10\mathrm{s}.$$

Answered by ajfour last updated on 20/Feb/18

$$maxrange=\frac{u^2}{g}=0.8mat\theta =45°$$$$\Rightarrow u=\sqrt{8}=2\sqrt{2}m/s$$$$TimeofflightT=\frac{2u\sin 45°}{g}$$$$T=0.4s$$$$Totaltime=(0.4s)\times no.ofjumps$$$$=0.4\times \frac{20}{0.8}=10s.$$

Commented by Tinkutara last updated on 23/Feb/18

Thank you very much Sir! I got the answer.

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