Question

A heavy particle is projected from a point on the horizontal at an angle 60 degree with the horizontal with a speed of 10 M per second then the radius of the curvature of its path at the instant of crossing the same horizontal is.

Answer 1

✔✔✅ Answer ✅✔✔

As the particle moves from its initial position to its final position, its vertical velocity decreases at the rate of 9.8 m/s each second. During this time its horizontal velocity is constant. Let’s determine the vertical and horizontal components of the initial velocity. 

Vertical = 10 x sin 60=5

√

3

, Horizontal = 10 x cos 60 = 5 m/s 

In a projectile horizontally components  of velocity remains constant.

...     10cos60°=Vcos30°...     V=10cos60°sec30°=10×

1

2

×

2

√

3

=

10

√

3

ans................... D)10/root3 m/s

☺☺☺