Question

1. A projectile cover double range as compare to its
maximum height attained. The angle of projection
from horizontal is-
(1) tan-2
(2) tan-14
(3) tan
(4) tan 5
-1
13​

Answer 1

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Answer 2

Answer:

$\boxed{\mathfrak{(1) \ tan^{-1}(2)}}$

Explanation:

Range of the projectile (R):

$\boxed{ \bold{R = \dfrac{v_0 ^{2} sin 2 \theta}{g} }}$

Maximum height of the projectile (H):

$\boxed{ \bold{H_{max} = \dfrac{v_0 ^{2} sin ^2 \theta}{2g}}}$

Where:

$\sf v_0 \rightarrow$ Initial velocity of the projectile

$\sf \theta \rightarrow$ Angle made by the projectile with horizontal

g $\sf \rightarrow$ Acceleration due to gravity

According to the question range of the projectile is double the maximum height of the projectile.

$\sf \implies R = 2H_{max} \\ \\ \sf \implies \dfrac{ \cancel{v_0 ^{2}} sin 2 \theta}{ \cancel{g}} = \cancel{2}(\dfrac{ \cancel{v_0 }^{2} sin ^{2} \theta}{ \cancel{ 2}\cancel{g}} ) \\ \\ \sf \implies 2 \cancel{sin \theta }.cos \theta = sin ^{ \cancel{2}} \theta \\ \\ \sf \implies 2cos \theta = sin \theta \\ \\ \sf \implies \frac{sin \theta}{cos \theta } = 2 \\ \\ \sf \implies tan \theta = 2 \\ \\ \sf \implies \theta = {tan}^{ - 1} (2)$

$\therefore$

Angle of projection from horizontal ($\sf \theta$) = $\sf tan^{-1}(2)$