# If a projectile is shot at an angle of pi/4 and at a velocity of 5 m/s, when will it reach its maximum height??

Jun 5, 2016

${t}_{e} = 0.36 \text{ } s$

#### Explanation:

$\alpha = \frac{\pi}{4}$

${v}_{i} = 5 \text{ } \frac{m}{s}$

${t}_{e} = {v}_{i} \cdot \sin \frac{\alpha}{g} \text{ time elapsed }$

${t}_{e} = \frac{5 \cdot 0.707}{9.81}$

${t}_{e} = 0.36 \text{ } s$