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Motion in a Plane — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane3 Marks
Question
The maximum height attained by a projectile is increased by 10% by increasing its speed of projection, without changing the angle of projection. What will the percentage increase in the horizontal range?
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Answer

As, maximum height, $\text{H}=\frac{\text{u}^2}{2\text{g}}\sin^2\theta$
Consider $\Delta\text{H}$ be the increase in H when u changes by $\Delta\text{u},$ it can be obtained by differentiating the above equation, we get
$\Delta\text{H}=\frac{2\text{u}\Delta\text{u}\sin^2\theta}{2\text{g}}=\frac{2\Delta\text{u}}{\text{u}}\text{H}$
$\Rightarrow\ \frac{\Delta\text{H}}{\text{H}}=\frac{2\Delta\text{u}}{\text{u}}$
Given, % increase in H is 10%, so
$\frac{\Delta\text{H}}{\text{H}}=\frac{10}{100}=0.1\Rightarrow\ \frac{2\Delta\text{u}}{\text{u}}=0.1$
As, $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$
$\because\ \Delta\text{R}=\frac{2\text{u}\Delta\text{u}}{\text{g}}\sin2\theta$
$\Rightarrow\ \frac{\Delta\text{R}}{\text{R}}=\frac{2\Delta\text{u}}{\text{u}}=0.1$
$\therefore$ % increase in horizontal range $=\frac{\Delta\text{R}}{\text{R}}\times100$
$=0.1\times100=10\%$

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