First slide

Projection Under uniform Acceleration

Question

At a height 0.4 m from the ground, the velocity of a projectile in vector form is $\vec{v} = (6 \hat{i} + 2 \hat{j}) {ms}^{- 1}$ The angle of projection is

Moderate

A

45^o
B

60^o
C

30^o
D

tan-1 (3/4)

Solution

$v^{2} = u^{2} - 2 gh or u^{2} = v^{2} + 2 gh u_{x}^{2} + u_{y}^{2} = v_{x}^{2} + v_{y}^{2} + 2 gh, u_{x} = v_{x} u_{y}^{2} = v_{y}^{2} + 2 gh or u_{y}^{2} = (2)^{2} + 2 \times 10 \times 0.4 = 12 u_{y} = \sqrt{12} = 2 \sqrt{3} {ms}^{- 1}, u_{x} = v_{x} = 6 {ms}^{- 1} tanθ = \frac{u_{y}}{u_{x}} = \frac{2 \sqrt{3}}{6} = \frac{1}{\sqrt{3}} \Rightarrow θ = 30 °$

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