Two bodies are thrown simultaneously from the same point. One thrown straight up and the other at an angle `alpha` with the horizontal. Both the bodies have velocity equal to `u`. Find the separation between the bodies at time `t`.

Explanation

After time `t`
Particle 1: `x_(1) = 0, y_(1) = ut - (1)/(2)g t^(2)`
Particle 2: `x_(2) = u cos alpha t, y_(2) = u sin alpha t - (1)/(2) g t^(2)`
Distance between the particles
`d = sqrt((x_(2) - x_(2))^(2) + (y_(2) - y_(1))^(2))`
`= sqrt((u cos alpha t - 0)^(2) + (u sin alpha t -u t)^(2))`
`= u tsqrt(cos^(2) alpha + sin^(2) alpha + 1 - 2 sin alpha)`
`= u t sqrt(2 - 2 sin alpha)`
`= sqrt(2) ut sqrt(1- sin alpha)` (i)
This answer can be expressed in other forms.
`sqrt(1 - sin alpha) = sqrt(sin^(2)(alpha)/(2) + cos^(2)(alpha)/(2) - 2 sin (alpha)/(2)cos(alpha)/(2))`
`= cos (alpha)/(2) - sin(alpha)/(2)`
`d = sqrt(2) u t(cos(alpha)/(2) - sin (alpha)/(2))` (ii)
`cos(alpha)/(2) - sin(alpha)/(2) = sqrt(2)((1)/(sqrt(2)) cos(alpha)/(2) - (1)/(sqrt(2)) sin(alpha)/(2))`
`= sqrt(2)( cos(pi)/(4) cos(alpha)/(2) - sin(pi)/(4) sin(alpha)/(2))`
`= sqrt(2) cos (9pi)/(4) + (alpha)/(2))`
`d = sqrt(2) ut sqrt(2) cos ((pi)/(4) + (alpha)/(2))`
`= 2 ut cos ((pi)/(4) + (alpha)/(2))` (iii)
`cos((pi)/(4) + (alpha)/(2)) = sin[(pi)/(2) - ((pi)/(4) + (alpha)/(2))]`
`= sin ((pi)/(4) + (alpha)/(2))`
`d = 2 ut sin ((pi)/(4) - (alpha)/(2))` (iv)
`[[cos (A + B) = cos A cos B - sin A sin B,],[cos (A - B) = cos A cos B + sin A sin B,]]`