# A projectile is shot from the ground at an angle of pi/8  and a speed of 4 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Mar 28, 2016

${x}_{m} = 1 , 15 \text{ m}$

#### Explanation:

${x}_{m} = \frac{{v}_{i}^{2} \cdot \sin 2 \alpha}{g}$
${v}_{i} = 4 \text{ } \frac{m}{s}$

alpha=pi/8;" "2*pi/8=pi/4 " "sin( pi/4)=sqrt2/2

${x}_{m} = \frac{{4}^{2} \cdot \frac{\sqrt{2}}{2}}{9 , 81}$

${x}_{m} = \frac{8 \sqrt{2}}{9 , 81}$

${x}_{m} = 1 , 15 \text{ m}$