A particle is moving along a vertical of radiusr=20 m with a constant vertical circle of radius r=20 m with a constnt speed v=31.4 m/s as shown in figure. Straight line ABC is horizontal and passes through the centre of the circle.A shell is fired from point A at the instant when the particle is at C. If distance AB is 20sqrt(3) m and the shell collide with the particle at B, then prove tan theta=((2n-1)^(2))/(sqrt(3)). Where n is an integer.Further show that smallest value of theta is 30^(@)

A particle is moving along a vertical of radiusr=20 m with a constant

1.	A particle is moving along a vertical of radiusr=20 m with a constant vertical circle of radius r=20 m with a constnt speed v=31.4 m/s as shown in figure. Straight line ABC is horizontal and passes through the centre of the circle.A shell is fired from point A at the instant when the particle is at C. If distance AB is 20sqrt(3) m and the shell collide with the particle at B, then prove tan theta=((2n-1)^(2))/(sqrt(3)). Where n is an integer.Further show that smallest value of theta is 30^(@)
Answer» Solution :As at the time of firing of the shell, the particle was at C and the shell collides with it at B, therefore thenumber of the revolutions completed by the particle is odd multiple of half i.e. `(2n-1)//2`, where n is an integer. Let T be the time period of the particle then `T=(2pir)/v=(2xx3.14xx20)/31.4=4` second If t be the time of the flight of the shell, then t=time of `[(2n-1)//2]` revolutions of the particle `=((2n-1))/2xx4=(2n-1)` second for a projectile the time of flight is GIVEN by `t=(2u sin theta)/g` Hence `(2usin theta)/g=(2(2n-1)`.......i The RANGE of the projectile is given by `R=(U^(2)sin 2 theta)/g` Hence `(u^(2) sin 2 theta)/g = 20 sqrt(3)` ......ii From equations i and ii `tan theta=((2n-1)^(2))/(sqrt(3))` For `theta` to be SMALLEST n=1 so `tan theta=((2n-1)^(2))/(sqrt(3))`