A body is projected at t = 0 with a velocity 10 ms-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 = 10 ms-2, the value of R is :

Question

A body is projected at t = 0 with a velocity 10 ms-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 = 10 ms-2, the value of R is : (A) 2.5 m (B) 10.3 m (C) 2.8 m (D) 5.1 m

Tardigrade Admin · Accepted Answer

vx = 10 cos 60° = 5 m/s vy = 10 cos 30° = 5 √3 m/s velocity after t = 1 sec. vx = 5 m/s vy = |( 5 √3 - 10)| m/s = 10 - 5 √3 an = (v2/R) ⇒ R = (v2x +v2y/an) = (25+100+75 -100√3/10 cosθ) tanθ = (10-5√3/5) = 2 -√3 ⇒ θ =15° R = (100(2-√3)/10 cos15) = 2.8 m

Q. 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=10 ms−2, the value of R is :

2.5 m

10.3 m

2.8 m

5.1 m

vx​=10cos60∘=5m/s vy​=10cos30∘=53​m/s velocity after t = 1 sec. vx​=5m/s vy​=∣(53​−10)∣m/s=10−53​ an​=Rv2​⇒R=an​vx2​+vy2​​=10cosθ25+100+75−1003​​ tanθ=510−53​​=2−3​⇒θ=15∘ R=10cos15100(2−3​)​=2.8m

Q. A body is projected at $t = 0$ with a velocity $10 m s^{- 1}$ at an angle of $6 0^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t = 1 s$ is $R$ . Neglecting air resistance and taking acceleration due to gravity $g = 10$ $m s^{- 2}$ , the value of $R$ is :