Equation of motion of a projectile is: y = 12x - frac{5}{9}{x^2} . horizontal component

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3-2.Motion in Plane

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The equation of motion of a projectile is: $y = 12x - \frac{5}{9}{x^2}$. The horizontal component of velocity is $3\ ms^{- 1}$ . Given that $g = 10\ ms^{- 2}$ , .......... $m$ is the range of the projectile .

A

$12.4$

B

$21.6$

C

$30.6$

D

$36.0$

Solution

$y=12 x-\frac{3}{4} x^{2}$

$\frac{d y}{d t}=12 \frac{d x}{d t}-\frac{3}{2} x \frac{d x}{d t}$

At $x=0: \quad \frac{d y}{d t}=12 \frac{d x}{d t}$

If $\theta$ be the angle of projection, then

$\frac{d y / d t}{d x / d t}=12=\tan \theta$

Also, if $u$ $=$initial velocity, then $u \cos \theta=3$

Hence, $\tan \theta \times u \cos \theta=36$ or $u \sin \theta=36$

Range, $R=\frac{u^{2} \sin 2 \theta}{g}=\frac{2 u^{2} \sin \theta \cos \theta}{g}$

$=\frac{2(u \sin \theta)(u \cos \theta)}{10}=\frac{2 \times 36 \times 3}{10}=21.6 \mathrm{m}$

Standard 11

Physics

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