Question

A particle is projected with a velocity of 30 ms^-1, at an angle of $θ_{0} = \tan^{- 1} (\frac{3}{4})$ . After 1 s, the particle is moving at an angle $θ$ to the horizontal, where tan $θ$ will be equal to (Take, g = 10 ms^-2 )

Moderate

Solution

$u_{x} = u \cos θ_{0} = 30 \times \frac{4}{5} = 24 {ms}^{- 1}$
$and u_{y} = u \sin θ_{0} = 30 \times \frac{3}{5} = 18 {ms}^{- 1}$
After 1 s, u_x will remain same and u_y will decrease by 10 ms^-1 or it will become 8 ms^-1.
$∴ \tan θ = \frac{v_{y}}{v_{x}} = \frac{8}{24} = \frac{1}{3}$

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