# A projectile is shot from the ground at an angle of pi/12  and a speed of 2/5 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Nov 18, 2017

#### Explanation:

Assume the projectile is launched with a speed ${v}_{0}$ at an angle $\setminus {\theta}_{0}$.

Time of Flight: $T = \frac{2 {v}_{0} \setminus \sin \setminus {\theta}_{0}}{g}$

Range of the projectile: $R = \frac{2 {v}_{0}^{2} \setminus \sin \setminus {\theta}_{0} \setminus \cos \setminus {\theta}_{0}}{g} = \frac{{v}_{0}^{2} \setminus \sin 2 \setminus {\theta}_{0}}{g}$

Height of the projectile: $H = \frac{{v}_{0}^{2} \setminus {\sin}^{2} \setminus {\theta}_{0}}{2 g}$

Given that : v_0 = 2/5 ms^{-1}; \qquad \theta_0 = \pi/12
Now substitute the numerical values of ${v}_{0}$, $\setminus {\theta}_{0}$ and $g$, calculate the numerical values of $R$ and $H$.

When the projectile is at maximum height, its coordinates are $\left(\frac{R}{2} , H\right)$

So its distance from the origin is ; $r = \setminus \sqrt{{\left(\frac{R}{2}\right)}^{2} + {H}^{2}} = \frac{\setminus \sqrt{{R}^{2} + 4 {H}^{2}}}{2}$

From the numerical values of $R$ and $H$, calculated earlier, evaluate $r$.