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Physics Question on Motion in a plane

A body is projected at such angle that the horizontal range is three times the greatest height. The angle of projection is

A

42∘8′42{^\circ} 8'

B

53∘7′53^{\circ} 7'

C

33∘7′33^{\circ} 7'

D

25∘8′25^{\circ} 8'

Answer

53∘7′53^{\circ} 7'

Explanation

Solution

Let a body be projected at a velocity uu at an angle θ\theta with the horizontal. Then horizontal range covered is given by


R=u2sin⁡2θg   ...R=\frac{u^{2} \sin 2 \theta}{g}\,\,\,...(i)
and height HH is
H=u2sin⁡2θ2g   ...H=\frac{u^{2} \sin ^{2} \theta}{2 g}\,\,\,...(ii)
Given R=3HR = 3H
∴u2 sin⁡ 2θg=3×u2 sin⁡2 θ2g\therefore \frac{ u^2 \, \sin \, 2 \theta }{ g } = 3 \times \frac{ u^2 \, \sin^2 \, \theta }{ 2 g }
Also, sin⁡2θ=2sin⁡θcos⁡θ\sin 2 \theta = 2 \sin \theta \cos \theta
∴u22 sin⁡θ cos⁡θg=3×u2 sin⁡2 θ2g\therefore \frac{ u^2 2 \, \sin \theta \, \cos \theta }{ g } = 3 \times \frac{ u^2 \, \sin^2 \, \theta }{ 2 g }
⇒2cos⁡θ=1.5sin⁡ θ\Rightarrow 2 \cos \theta = 1.5 \sin \, \theta
⇒tan⁡ θ=21.5=1.33\Rightarrow \tan \, \theta = \frac{ 2}{ 1.5 } = 1.33
⇒θ=53∘ 7′\Rightarrow \theta = 53^\circ \, 7'
Hence, angle of projection is 53∘ 7′53^\circ \, 7'