A ball is projected with a speed of 2ms^ – 1 with respect to a...

A ball is projected with a speed of

2ms^{– 1}

with respect to a trolley which is moving with

4ms^{– 1}

, as shown in the figure. The angle at which ball is projected with respect to the horizontal in ground frame is

A ball is projected with a speed of 2ms^ – 1 with respect to a trolley which is moving with 4ms^ – 1 , as shown in the

A ball is projected with a speed of 2ms^ – 1 with respect to a trolley which is moving with 4ms^ – 1 , as shown in the

Options

A $\tan^{- 1} \left(\sqrt{\dfrac{3}{28}}\right)$
B $\tan^{- 1} \left(\dfrac{5}{\sqrt{3}}\right)$
C $\sin^{- 1} \left(\sqrt{\dfrac{3}{28}}\right)$
D $\sin^{- 1} \left(\dfrac{5}{\sqrt{28}}\right)$

Correct answer

C. $\sin^{- 1} \left(\sqrt{\dfrac{3}{28}}\right)$

Step-by-step solution

{\overset{→}{v}}_{b} = {\overset{→}{v}}_{bT} + {\overset{→}{v}}_{T}

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tan ϕ = \dfrac{2sin {60}^{o}}{4 + 2cos {60}^{o}} = \dfrac{\sqrt{3}}{5}

<br /><br />

sin ϕ = \sqrt{\dfrac{3}{28}}

<br /><br />

ϕ = \sin^{- 1} \left(\sqrt{\dfrac{3}{28}}\right)

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