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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2023 · Paper 2 · Physics · Motion in Two Dimensions

A projectile is projected at 30^ ^ from horizontal with initial...

A projectile is projected at 30∘30^{{}^{\circ}} from horizontal with initial velocity 40ms−140 \mathrm{ ms}^{-1}. The velocity of the projectile at t=2s\mathrm{t}=2 \mathrm{ s} from the start will be: (Given g=10m/s2\mathrm{g}=10 \mathrm{ m} / \mathrm{s}^2 )

Options

  1. A203ms−120 \sqrt{3} \mathrm{ ms}^{-1}
  2. B403ms−140 \sqrt{3} \mathrm{ ms}^{-1}
  3. C20ms−120 \mathrm{ ms}^{-1}
  4. D
    Zero

Correct answer

A. 203ms−120 \sqrt{3} \mathrm{ ms}^{-1}

Step-by-step solution

Given,
Initial velocity of projectile, u=40m/su=40 \mathrm{ m} / \mathrm{s}
Angle, θ=30∘\theta=30^{{}^{\circ}}
Time of flight
\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{2 \times 40 \times 1}{10 \times 2}=45 \left(\because \mathrm{g}=10 \mathrm{ m}/ \mathrm{s}^2\right)<br />It means projectile is at maximum height at t=2 s.Atmaximumheightverticalcomponent\mathrm{ s}.At maximum height vertical component of velocityiszero.<br/>velocity is zero.<br /> Velocity at t=2s=Vx=ucos⁡θ=40cos⁡30∘=203ms−1. \begin{aligned}\\ \text{ Velocity at } \mathrm{t}=2 \mathrm{ s}=\mathrm{V}_{\mathrm{x}}=\mathrm{u} \cos \theta=40 \cos 30^{{}^{\circ}} \\ \\ & =20 \sqrt{3} \mathrm{ ms}^{-1} \text{. }\\ \\ & \\ \end{aligned}$$
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