# 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?

Time to reach at highest point $t = \frac{u \sin \alpha}{g}$
So horizontal displacement during this period ${D}_{h} = u \cos \alpha \cdot t = \frac{{u}^{2} \sin \alpha \cos \alpha}{g} = \frac{1}{2} \frac{{u}^{2} \sin 2 \alpha}{g}$
Vertical maximum height covered during this time ${D}_{v} = \frac{{u}^{2} {\sin}^{2} \alpha}{2 g}$
so distance from the point of projection $D = \sqrt{{D}_{h}^{2} + {D}_{v}^{2}}$