Consider a pendulum consisting of a massless string with a mass at one end. The mass is held with the string horizontal and then released. The mass swings down, and on its way back up, the string is cut at point P when it makes an angle of with `theta` the vertical. Find the angle for `theta` which horizontal range .R. is maximum.

Explanation

The speed at angle is given by conservation of energy.

`(1)/(2)mv^(2)=mgh rArr (1)/(2) mv^(2)=mgl cos theta`
`v = sqrt(2gl cos theta)`
When string is cut particle moves as a projectile with velocity component.
`v_(x)= v cos theta v_(y)=v sin theta`
The time of flight is `t=2((v_(y))/(g))`
Range is, `R = v_(x)t = v_(x)((2v_(y))/(g))`
`= (2v_(x)v_(y))/(g)=(2(v cos theta)(v sin theta))/(g)=(2 v^(2)sin theta cos theta)/(g)`
`= (2 (2gl cos theta)sin theta cos theta)/(g)= 4l cos^(2) theta sin theta`
Differentiating .R. with respect to `.theta.` and equating it to zero we get
`0=4l (-2 cos theta sin theta)sin theta + cos^(2)theta cos theta`
or, `2 sin^(2)theta cos theta = cos^(3)theta`, or ` tan theta = 1//sqrt(2)`