Question

The velocity at the highest point of a projectile projected at 60° with the horizontal with velocity v is: (a) o b) v/2 (c) v /4 (d) v /6​

Answer 1

Explanation:

The radius of curvature will be minimum at the highest point of trajectory.

Horizontal velocity at that point v

h

=10cosθ=10×0.5=5 m/s

Vertical velocity at that point v

v

=0

⇒m

R

min

v

h

2

=mg

⇒R

min

=

g

v

h

2

=

9.8

(5)

2

≃2.55m