The maximum height attained by projectile is increased by $10 \%$ by keeping the angle of projection constant. What is the percentage increase in the time of flight?

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext {} \mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}}, \mathrm{~T}=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{~g}} \ & \Rightarrow \frac{\mathrm{H}}{\mathrm{~T}}=\frac{\mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}} 	imes \frac{\mathrm{g}}{2 \mathrm{u} \sin 	heta}=\frac{1}{4} \mathrm{u} \sin 	heta \end{aligned}$ Also, $\mathrm{u} \mu \mathrm{T}$ $\begin{aligned} & 	herefore \frac{\mathrm{H}}{\mathrm{~T}} \propto \mathrm{~T} \Rightarrow \mathrm{H} \propto \mathrm{~T}^2 \Rightarrow\left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}ight)=\sqrt{\frac{1.1 \mathrm{H}_1}{\mathrm{H}_1}}=\sqrt{1.1} \ & 	herefore \mathrm{~T}_2=\sqrt{1.1} \mathrm{~T}_1=1.05 \mathrm{~T}_1 \ & 	herefore \Delta \mathrm{~T} \%=5 \% \end{aligned}$

The maximum height attained by projectile is increased by $10 \%$ by keeping the angle of projection…

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