Question

A body is projected with a velocity `10m/s` at an inclination `45^o` to the horizontal. The minimum radius of curvature of the trajectory described by the particle is?

Answer 1

`2.54m`

Explanation:

As we know that projectile motion follows a parabolic pathway,so we can discuss about the properties of parabola to find the radius of curvature of the projectile motion.

Well the parabola is relatively flat on its two sides in comparison to its vertex.So we can consider sides of parabola to be part of very bigger circles that means,those parts have a greater radius of curvature.

On contrary to that,vertex of the parabola is the least flat,and that small amount of arc can be considered as a part of a circle having smaller radius of curvature.

Now,for a projectile motion,its vertex lies at the highest point of its motion,so we can say the height from ground or in other words,maximum height reached by a projectile is the minimum radius of curvature depicted by its trajectory.

In the given case applying the formula to find maximum height reached by the projectile,we get, `H =(u^2 sin^2 theta)/(2g)=100/(2×2×9.8)=2.54 m` (where, `u` is the velocity of projection and `theta` is the angle of projection w.r.t horizontal)

Answer 2

The radius of curvature is `=5.1m`

Explanation:

The equation of the trajectory of the projectile is

`y=xtantheta-(gx^2)/(2u^2cos^2theta)`

`u=10ms^-1`

`theta=45^@`

Therefore,

`y=x-(gx^2)/(200*1/2)=x-(gx^2)/100`

The radius of curvature is givenby the equation

`r=(1+(dy/dx)^2)^(3/2)/(|(d^2y)/dx^2|)`

`dy/dx=1-(2gx)/(100)=1-(gx)/50`

`(d^2y)/dx^2=-g/50`

So,

`r=(1+(1-(gx)/50)^2)^(3/2)/(g/50)`

`r=50/g*(1+(1-(gx)/50)^2)^(3/2)`

The minimum value is when

`(dr)/dx=0`

`(dr)/dx=-50/g*3/2*(1+(1-(gx)/50)^2)^(1/2)*2(1-(gx)/50)*g/50`

`=-3*(1+(1-(gx)/50)^2)*(1-(gx)/50)`

Therefore,

`(dr)/dx=0`, `=>`, `1-(gx)/50=0`

`x=50/g`

The minimum radius of curvature is

`r=50/g`

The acceleration due to gravity is `g=9.8ms^-2`

`r=50/9.8=5.1m`