JEE Mains · Physics · STD 11 - 3.2 motion in plane
The range of the projectile projected at an angle of \(15^{\circ}\) with horizontal is \(50\,m\). If the projectile is projected with same velocity at an angle of \(45^{\circ}\) with horizontal, then its range will be \(........\,m\)
- A \(50\)
- B \(50 \sqrt{2}\)
- C \(100\)
- D \(100 \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(100\)
Step-by-step Solution
Detailed explanation
\(R =\frac{ v ^2 \sin 2 \theta}{ g }\) \(R \propto \sin (2 \theta)\) \(\frac{ R _1}{ R _2}=\frac{\sin \left(2 \theta_1\right)}{\sin \left(2 \theta_2\right)}=\frac{\sin (2 \times 15)}{\sin (2 \times 45)}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}}\) \(\frac{50}{ R _2}=\frac{1}{2}\)…
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