JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
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
The range of a projectile is given by: $$R = \frac{u^2 \sin 2\theta}{g}$$
At $$\theta = 15°$$: $$R_1 = \frac{u^2 \sin 30°}{g} = \frac{u^2}{2g} = 50$$ m
At $$\theta = 45°$$: $$R_2 = \frac{u^2 \sin 90°}{g} = \frac{u^2}{g}$$
From the first equation: $$\frac{u^2}{g} = 100$$ m
Therefore $$R_2 = 100$$ m.
The correct answer is Option 1: 100 m.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
