Q81 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics
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 :
- A50 m
- B$50 \sqrt{2} \mathrm{~m}$
- Correct answerC100 m
- D$100 \sqrt{2} \mathrm{~m}$
Solution
Range of projectile
$R=\frac{v^2 \sin 2 \theta}{g}(\because R \propto \sin (2 \theta))$
$\frac{R_1}{R_2}=\frac{\sin \left(2 \theta_1\right)}{\sin \left(2 \theta_2\right)}=\frac{\sin (2 \times 15)}{\sin (2 \times 45)}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}}$
$\Rightarrow \frac{50}{\mathrm{R}_2}=\frac{1}{2} \Rightarrow \mathrm{R}_2=100 \mathrm{~m}$
Hence, the correct answer is option: 'C'.