Question bank/JEE Main/Physics
JEE MainPhysicsMedium
A ball was thrown at velocity of 10 m/s at 37° to the horizontal.
What is the rate of change of its speed when its radius of curvature is
51.2 m.
Show answer and solution
[IMAGE] aN = g cos θ =[IMAGE] v cos θ = 10 cos 37°
v =[IMAGE] g cos3 θ =[IMAGE] cos3 θ =[IMAGE] cos θ =[IMAGE] θ = 60°
at =[IMAGE] = g sin θ = 2
TOPIC: CIRCULAR MOTION
SUB TOPIC: Radius of Curvature
v =[IMAGE] g cos3 θ =[IMAGE] cos3 θ =[IMAGE] cos θ =[IMAGE] θ = 60°
at =[IMAGE] = g sin θ = 2
TOPIC: CIRCULAR MOTION
SUB TOPIC: Radius of Curvature
Source: Grass-G / P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx
Need a video for this idea?
Scan any doubt from your notes and get a whiteboard explanation.