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Question:

A body is projected at t=0 with a velocity 10ms⁻¹ at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t=1s is R. Neglecting air resistance and taking acceleration due to gravity g=10ms⁻², the value of R is:

10.3m

2.4m

2.8m

5.1m

Solution:

vx=10cos60°=5m/s
vy=10cos30°=5√3m/s
Velocity after t=1sec
vx=5m/s
vy=|(5√3−10)|m/s=10−5√3
an=v²/R ⇒vx²+vy²
an=25+(10−5√3)²
10cosθtanθ=10−5√3/5=2−√3 ⇒θ=15°
R=100/(2−√3)10cos15=2.8m