JEE Mains · Physics · STD 11 - 3.2 motion in plane
Two projectiles are projected with the same initial velocities at the \(15^\circ\) and \(30^\circ\) with respect to the horizontal. The ratio of their ranges is \(1:x\). The value of \(x\) is:
- A \(\sqrt{2}\)
- B \(\sqrt{3}\)
- C \(2\sqrt{3}\)
- D \(\dfrac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
The horizontal range of a projectile is given by the formula \(R = \dfrac{u^2 \sin(2\theta)}{g}\). For the first projectile, the angle of projection is \(\theta_1 = 15^\circ\). \(R_1 = \dfrac{u^2 \sin(2 \times 15^\circ)}{g} = \dfrac{u^2 \sin(30^\circ)}{g} = \dfrac{u^2}{2g}\) For…
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