Question

A projectile is shot from the ground at an angle of `pi/12` and a speed of `6 m/s`. When the projectile is at its maximum height, what will its distance from the starting point be?

Answer 1

Time to reach at highest point `t = (usinalpha)/g`
So horizontal displacement during this period `D_h=ucosalpha*t=(u^2sinalpha cos alpha)/g=1/2(u^2sin2alpha)/g`

Vertical maximum height covered during this time `D_v=(u^2sin^2alpha)/(2g)`
so distance from the point of projection `D=sqrt(D_h^2+D_v^2)`
please put the values and calculate