First slide

Projection Under uniform Acceleration

Question

The equations of motion of a projectile are given by x = 36 t metre and $2 y = 96 t - 9 \cdot 8 t^{2}$ metre. The angle of projection is

Moderate

A

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

$\sin^{- 1} (\frac{3}{5})$
C

$\sin^{- 1} (\frac{4}{3})$
D

$\sin^{- 1} (\frac{3}{4})$

Solution

Given that
x = 36 t and $2 y = 96 t - 9 \cdot 8 t^{2}$
From the given equations, we have
$V_{x} = \frac{dx}{dt} = 36$ and $v_{y} = \frac{dy}{dt} = 48 - 9 \cdot 8 t$
At $t = 0, V_{y} = 48$
$∴ V = \sqrt{(V_{x}^{2} + V_{y}^{2})} = 60$
This is shown in fig. (8).
From figure
$\sin θ = \frac{V_{y}}{V} = \frac{48}{60} = \frac{4}{5}$
or $θ = \sin^{- 1} (4 / 5)$

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