Question

A particle is projected with a velocity 30 ms-lat an angle of 60°with the horizontal. The vertical component of velocity at its
maximum height is
(A) 15 ms
(B) 15 3 ms-1
(C) 20 ms-1
(D) Zero​

Answer 1

Explanation:

We have,,

Initial speed of particle, u = 30 m/s

Angle of projection, ∅ = 60°

Acc due to gravity, a = -g = -9.8 m/s

At the maximum height,

(Consider the motion in y-direction,,)

We have,

$\sf{u_{y}=u\sin(\Theta)}$

$\implies{\sf{u_{y}=30\times{\sin(60°)}}}$

$\implies{\sf{u_{y}=30\times{\frac{\sqrt{3}}{2}}}}$

$\implies{\boxed{\sf{u_{y}=+15\sqrt{3} \:ms^{-1}}}}$

Max height attained,

$\sf{H=\frac{u^{2}\sin^{2}(\Theta)}{2g}}$

$\implies{\sf{H=\frac{(30)^{2}\sin^{2}(60°)}{2\times{9.8}}}}$

$\implies{\sf{H=\frac{900\times{(\frac{\sqrt{3}}{2})^{2}}}{19.6}}}$

$\implies{\sf{H=\frac{900\times{\frac{3}{4}}}{19.6}}}$

$\implies{\sf{H=\boxed{\sf{\frac{675}{19.6}\:m}}}}$

Therefore,

Vertical velocity of particle at maximum height is given by,

$\boxed{\sf{v_{y}^{2}-u_{y}^{2}=2aH}}$

$\implies{\sf{v_{y}^{2}-(15\sqrt{3})^{2}=2\times{(-9.8)}\times(\frac{675}{19.6})}}$

$\implies{\sf{v_{y}^{2}-675=2\times(\frac{-675}{2})}}$

$\implies{\sf{v_{y}^{2}=-675+675}}$

$\implies{\sf{v_{y}=\sqrt{0}}}$

$\implies{\underline{\boxed{\sf{v_{y}=0\:ms^{-1}}}}}$

Therefore option (D) is correct...✓✓✓

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Answer 2

Answer:

The answer is a option A

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