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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2015 · Physics · Motion in Two Dimensions

The position of a projectile launched from the origin at t=0 is...

The position of a projectile launched from the origin at t=0t=0 is given by r⃗=(40i^+50j^)m\vec{r}=(40 \hat{i}+50 \hat{j}) \mathrm{m} at t=2 st=2 \mathrm{~s}. If the projectile was launched at an angle θ\theta from the horizontal, then θ\theta is \left(\right.takeg=10 ms−2)take \left.g=10 \mathrm{~ms}^{-2}\right)\

Options

  1. Atan⁡−123\tan ^{-1} \frac{2}{3}
  2. Btan⁡−132\tan ^{-1} \frac{3}{2}
  3. Ctan⁡−174\tan ^{-1} \frac{7}{4}
  4. Dtan⁡−145\tan ^{-1} \frac{4}{5}

Correct answer

C. tan⁡−174\tan ^{-1} \frac{7}{4}

Step-by-step solution

From question, Horizontal velocity (initial), u_{x}=402=20m\frac{40}{2}=20 \mathrm{ m}/ s\mathrm{s} Vertical velocity (initial), 50=uyt+12gt250=u_{y} t+\frac{1}{2} g t^{2}⇒uy×2+12(−10)×4 or, 50=2uy−20\begin{array}{l} \Rightarrow u_{y} \times₂⁺\frac{1}{2}(-10) \times4 \\ \text{ or, } \quad50=2 u_{y}-20 \end{array}or,or, u_{y}=702=35m\frac{70}{2}=35 \mathrm{ m} / s\mathrm{s}∴tan⁡θ=uyux=3520=74⇒ Angle θ=tan⁡−174\begin{array}{l} \therefore \quad \tan \theta=\frac{u_{y}}{u_{x}}=\frac{35}{20}=\frac{7}{4} \\ \Rightarrow \quad \text{ Angle } \theta=\tan ^{-1}\frac{7}{4} \end{array}$
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