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JEE Main 2003 (AIEEE) · Projectile Basics & Trajectory

Projectile Basics & Trajectory question from JEE Main 2003 (AIEEE)

Q10JEE Main 2003 · AIEEEMedium

Projectile Basics & Trajectory

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30∘30^\circ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? [g=10 m/s2, sin⁡30∘=12, cos⁡30∘=32]\left[g = 10\text{ m/s}^{2},\ \sin 30^\circ = \frac{1}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2}\right]

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Answer: (d)

  1. Height 10 m from ground is the level of the throwing point, so we need the horizontal range on that level.
  2. R=u2sin⁡2θg=100sin⁡60∘10=10×32R = \frac{u^2\sin2\theta}{g} = \frac{100\sin60^\circ}{10} = 10\times\frac{\sqrt3}{2}
  3. R=8.66R = 8.66 m
  4. So the answer is (d) 8.66 m.
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