Question

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

Moderate

Solution

It is given that, x = 36 t
$∴ v_{x} = \frac{d x}{d t} = 36 {ms}^{- 1} and y = 48 t - 4.9 t^{2}$
$∴ v_{y} = 48 - 9.8 t$
$At t = 0, v_{x} = 36 {ms}^{- 1} and v_{y} = 48 {ms}^{- 1}$
So, angle of projection, $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})$
$= \tan^{- 1} (\frac{48}{36}) = \tan^{- 1} (\frac{4}{3})$
or $θ = \sin^{- 1} (\frac{4}{5})$

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