The maximum height attained by a projectile i | vedclass

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(A) The maximum height $H$ and time of flight $T$ of a projectile are given by:$H = \frac{u^2 \sin^2 	heta}{2g}$ and $T = \frac{2u \sin 	heta}{g}$.F

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$5$

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(A) The maximum height $H$ and time of flight $T$ of a projectile are given by:$H = \frac{u^2 \sin^2 	heta}{2g}$ and $T = \frac{2u \sin 	heta}{g}$.From these equations,we can see that $H \propto u^2$ and $T \propto u$ (since $	heta$ is constant).Therefore,$H \propto T^2$.If the height increases by $10 \%$,the new height $H_2 = 1.10 H_1$.Since $H \propto T^2$,we have $\frac{H_2}{H_1} = \left(\frac{T_2}{T_1}ight)^2$.$1.10 = \left(\frac{T_2}{T_1}ight)^2 \implies \frac{T_2}{T_1} = \sqrt{1.10} \approx 1.0488$.This corresponds to an increase of approximately $4.88 \%$,which is closest to $5 \%$.

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

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