# If a projectile is shot at an angle of pi/6 and at a velocity of 23 m/s, when will it reach its maximum height??

Apr 12, 2017

elapsed time to the maximum height is t=1.17 sec.

#### Explanation:

The initial velocity $\text{ " v_i" }$can be divided into two part as vertical
and horizontal.

• The part of $\text{ "v_x" }$supplies the horizontal movement.
• The part of $\text{ "v_y" }$ supplies the vertical movement.

• ${v}_{y} = {v}_{i} \cdot \sin \theta - g \cdot t$

• The $\text{ "v_y" }$part of the $\text{ } {v}_{i}$ decreases depending on the time.
• Please notice that the $\text{ "v_y" }$ is zero at the maximum height.
• if we write as ${v}_{y} = {v}_{i} . \sin \theta - g \cdot t = 0$
• We get ${v}_{i} . \sin \theta = g . t$
• if we solve according to t,We get $t = \frac{{v}_{i} \cdot \sin \theta}{g}$

• $t = \frac{23. \sin \left(\frac{\pi}{6}\right)}{9.81}$

• t=1.17 sec