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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2011 · Physics · Motion in Two Dimensions

A passenger in a open car travelling at 30 m / s throws a ball out...

A passenger in a open car travelling at 30m/s30 \mathrm{ m} / \mathrm{s} throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is 20m/s20 \mathrm{ m} / \mathrm{s} at 60∘60^{{}^{\circ}} to the horizontal. The angle of projection of the ball with respect to the horizontal road will be

Options

  1. Atan⁡−1(23)\tan ^{-1} \left(\frac{2}{3}\right)
  2. Btan⁡−1(34)\tan ^{-1} \left(\frac{\sqrt{3}}{4}\right)
  3. Ctan⁡−1(43)\tan ^{-1} \left(\frac{4}{\sqrt{3}}\right)
  4. Dtan⁡−1(34)\tan ^{-1} \left(\frac{3}{4}\right)

Correct answer

B. tan⁡−1(34)\tan ^{-1} \left(\frac{\sqrt{3}}{4}\right)

Step-by-step solution

(v→bc)x=(v→b)x−(v→c)x20cos⁡60∘=(v→b)x−30(v→b)x=40;(v→bc)y=(v→b)y−(v→c)y20sin⁡60∘=(v→b)y−0(v→b)y=103;tan⁡θ=(v→b)y(v→b)x=10340=34\begin{array}{l}\left(\overrightarrow{\mathrm{v}}_{\mathrm{bc}}\right)_{\mathrm{x}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}-\left(\overrightarrow{\mathrm{v}}_{\mathrm{c}}\right)_{\mathrm{x}}\\ 20 \cos 60^{{}^{\circ}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}-30 \\ \left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}=40 ;\left(\overrightarrow{\mathrm{v}}_{\mathrm{bc}}\right)_{\mathrm{y}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}-\left(\overrightarrow{\mathrm{v}}_{\mathrm{c}}\right)_{\mathrm{y}}\\ 20 \sin 60^{{}^{\circ}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}-0 \\ \left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}=10 \sqrt{3}; \tan \theta=\frac{\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}}{\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}}=\frac{10 \sqrt{3}}{40}=\frac{\sqrt{3}}{4} \end{array}$$
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