Exams › JEE Main › Physics
A stone is projected horizontally with a speed of 10 m/s. Find the radius of curvature of its path 3 s after the throw. (g = 10 m/s²)
- 10*sqrt(10) m
- 100*sqrt(10) m
- sqrt(10) m
- 100 m
Correct answer: 100*sqrt(10) m
Solution
At t = 3 s, vx = 10, vy = 30, v = sqrt(10² + 30²) = sqrt(1000) = 10*sqrt(10). a_perp = g*vx/v = 10*10/(10*sqrt(10)) = sqrt(10). R = v²/a_perp = 1000/sqrt(10) = 100*sqrt(10) m.
Related JEE Main Physics questions
⚔️ Practice JEE Main Physics free + battle 1v1 →