Question

The equations of motion of a projectile are given by x = 36t m and 2y = 96t – 9.8t² m. The angle of projection is

• A. $\sin^{- 1}\left( \frac{4}{5} \right)$
• B. $\sin^{- 1}\left( \frac{3}{5} \right)$
• C. $\sin^{- 1}\left( \frac{4}{3} \right)$
• D. $\sin^{- 1}\left( \frac{3}{4} \right)$

Answer 1

Answer: (A)

$\sin^{- 1}\left( \frac{4}{5} \right)$

Answer 2

Given: x = 36 t

and $2y = 96t - 9.8t^{2}$

or $y = 48t - 4.9t^{2}$

Let the initial velocity of projectile be u and angle of projections is $\theta$. Then,

Initial horizontal component of velocity.

$u_{x} = u\cos\theta = \left( \frac{dx}{dt} \right)_{t = 0} = 36$ …… (i)

Or u $\cos\theta = 36$

Initial vertical component of velocity

$u_{y} = u\sin\theta = \left( \frac{dy}{dt} \right)_{t = 0} = 48$

Or $u\sin\theta = 48$ ….. (ii)

$\tan\theta = \frac{48}{36} = \frac{4}{3}$

$\therefore\sin\theta = \frac{4}{3}or\theta = \sin^{- 1}\left( \frac{4}{5} \right)$