# If a projectile is shot at an angle of (7pi)/12 and at a velocity of 1 m/s, when will it reach its maximum height?

Mar 14, 2016

${t}_{e} = 0 , 098 s$
$\alpha = 7 \frac{\pi}{12}$
${v}_{i} = 1 \frac{m}{s}$
${t}_{e} = {v}_{i} \cdot \sin \frac{\alpha}{g} \text{ elapsed time to the maximum height}$
${t}_{e} = \frac{1 \cdot \sin \left(7 \frac{\pi}{12}\right)}{9 , 81}$
${t}_{e} = 0 , 098 s$