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JEE Main 2014 (9 Apr, Shift 1) · Projectile Basics & Trajectory

Projectile Basics & Trajectory question from JEE Main 2014 (9 Apr, Shift 1)

Q6JEE Main 2014 · 9 APR · SHIFT 1Medium

Projectile Basics & Trajectory

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

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Answer: (c)

  1. x=uxt⇒40=2ux⇒ux=20x = u_x t \Rightarrow 40 = 2u_x \Rightarrow u_x = 20 m/s
  2. y=uyt−12gt2⇒50=2uy−20⇒uy=35y = u_y t - \frac12gt^2 \Rightarrow 50 = 2u_y - 20 \Rightarrow u_y = 35 m/s
  3. tan⁡θ=uyux=3520=74\tan\theta = \frac{u_y}{u_x} = \frac{35}{20} = \frac74
  4. So the answer is (c) tan⁡−174\tan^{-1}\frac74.
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