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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Motion in Two Dimensions

BITSAT 2024 · Paper 2 · Physics · Motion in Two Dimensions

The range of the projectile projected at an angle of 15^ ^ with...

The range of the projectile projected at an angle of 15∘15^{{}^{\circ}} with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45∘45^{{}^{\circ}} with horizontal, then its range will be :

Options

  1. A50 m
  2. B502m50 \sqrt{2} \mathrm{ m}
  3. C100 m
  4. D1002m100 \sqrt{2} \mathrm{ m}

Correct answer

C. 100 m

Step-by-step solution

Range of projectile R=v2sin⁡2θg(∵R∝sin⁡(2θ))R1R2=sin⁡(2θ1)sin⁡(2θ2)=sin⁡(2×15)sin⁡(2×45)=sin⁡30∘sin⁡90∘⇒50R2=12⇒R2=100mR=\frac{v^2 \sin 2 \theta}{g}(\because 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}}}\Rightarrow \frac{50}{\mathrm{R}_2}=\frac{1}{2} \Rightarrow \mathrm{R}_2=100 \mathrm{ m}
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