Range projectile projected |ElectromagneticWavesJEEMain2023.

JEE MAIN 2023

10-4-2023

Question

The range of the projectile projected at an angle of 15° with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45º with horizontal, then its range will be:

Select the correct option:

50 m

100 m

$100 \sqrt{2} \mathrm{~m}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$ \left.R=\frac{v^2 \sin 2 \theta}{g} \right\rvert\, $$ $\mathrm{R} \propto \sin (2 \theta)$ $$ \begin{aligned} & \frac{R_1}{R_2}=\frac{\sin \left(2 \theta_1\right)}{\sin \left(2 \theta_2\right)}=\frac{\sin (2 \times 15)}{\sin (2 \times 45)}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}} \\ & \frac{50}{R_2}=\frac{1}{2} \\ & R_2=100 \mathrm{~m} \end{aligned} $$

Question Tags

JEE Main

Physics

Easy

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