Question

A body is projected at t = 0 with a velocity 10 ms⁻¹ at an angle of 60º 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 radius of R is:
(A) 10.3 m (B) 2.8 m
(C) 2.5 m (D) 5.1 m

Answer 1

vx=10cos60o=5m/s

vy=10cos30o=53m/s

Velocity after t=1 sec

vx=5m/s

vy=∣(53−10)∣m/s=10−53

an=Rv2⇒anvx2+vy2=10cosθ25+100+75−1003

tanθ=10−5root3=2−root ⇒θ=15o

R=100(2−root3)/10cos15=2.8m

Answer 2

v  

x

​  

=10cos60  

o

=5m/s

v  

y

​  

=10cos30  

o

=5  

3

​  

m/s

Velocity after t=1 sec

v  

x

​  

=5m/s

v  

y

​  

=∣(5  

3

​  

−10)∣m/s=10−5  

3

​  

 

a  

n

​  

=  

R

v  

2

 

​  

⇒  

a  

n

​  

 

v  

x

2

​  

+v  

y

2

​  

 

​  

=  

10cosθ

25+100+75−100  

3

​  

 

​  

 

tanθ=  

5

10−5  

3

​  

 

​  

=2−  

3

​  

⇒θ=15  

o

 

R=  

10cos15

100(2−  

3

​  

)

​  

=2.8m