Question

Projectile is a particle which is projected into air with an initial velocity against gravity.

A. What is the angle of projection for maximum horizontal range?

B. Obtain the expression for time of flight.

C. An insect is moving in a circle with uniform speed. Obtain the expression for centripetal acceleration (ac) in terms of angular speed (ω)​

Answer 1

Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .

The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.

Horizontal range OA = horizontal component of velocity × total flight time

      R=ucosθ*(2usinθ/g)

        R=u²sin2θ/g

SO for maximum horizontal range sin2θ=1

so we get, maximum horizontal range R=u²/g.

(B) The total time of flight is

Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

s=ut-1/2gt²

gt=2usinθ

t=2usinθ/g

(C)the direction of the centripetal acceleration is always inwards along the radius vector of the circular motion.

Now we know that, the centripetal force required to keep an insect of mass ‘m’ revolving in a circular path of radius ‘r’ with a speed ‘v’ is,  

F = mv²/r

so now centripetal acceleration is

a=F/m

a=(mv²/r)÷(m)=v²/r

now v=rω

so putting the value of v we get, a=r²ω²/r= ω²r ANSWER