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Model Paper 6 — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 65 Marks
Question
i. Show that for two complementary angles of projection of a projectile thrown with the same velocity, the horizontal ranges are equal.
ii. For what angle of projection of a projectile, is the range maximum?
iii. For what angle of projection of a projectile, are the horizontal range and maximum height attained by the projectile equal?
✓

Answer

i. Range of projectile $=\frac{u^2 \sin 2 \theta}{g}$
for angle $\theta$, Range $R _1=\frac{u^2 \sin 2 \theta}{g}$
for angle $(90-\theta)$, Range $R _2=\frac{u^2 \sin 2(90-\theta)}{g}=\frac{u^2 \sin (180-2 \theta)}{g} R _2=\frac{u^2 \sin 2 \theta}{g}$
therefore $R_1=R_2$ i.e., two complementary angle of projection of projectile thrown with the same velocity, the horizontal range is equal.
ii. Let $\theta$ be the angles of projection for projectile thrown with speed v
Range of projectile $=\frac{u^2 \sin 2 \theta}{g}$
for range to be maximum, $\sin 2 \theta=1$
or $2 \theta=90^{\circ}$
$\theta=45^{\circ}$
iii. Range of projectile $=\frac{u^2 \sin 2 \theta}{g}$
height of projectile $h =\frac{u^2 \sin ^2 \theta}{2 g}$
as Range = height of projectile
there fore, $\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin ^2 \theta}{2 g}$
$\begin{aligned} & 2 \sin \theta \cos \theta=\frac{\sin ^2 \theta}{2} \\ & \tan \theta=4 \\ & \theta=\tan ^{-1} 4\end{aligned}$

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