# A projectile is shot at a velocity of 4 m/s and an angle of pi/12 . What is the projectile's maximum height?

Jan 14, 2016

$h = \frac{{\left(4 \times 0.25882\right)}^{2}}{2 \times 9.81} m$
$\approx 0.05463 m$

#### Explanation:

Maximum height is reached because of $S \in \theta$ component of the velocity.
Use the formula
${v}^{2} - {u}^{2} = 2 g h$
Given $v = 0 , u = 4 . \sin \left(\frac{\pi}{12}\right)$; as $\theta = \frac{\pi}{12}$
Take $g = 9.81 \frac{m}{s} ^ 2$
Since gravity is acting against the velocity so it is deceleration and $-$ sign is used in front of g
Substituting various values
$h = \frac{{\left(4 . \sin \left(\frac{\pi}{12}\right)\right)}^{2}}{2 \times 9.81} m$
Use tables to substitute the value of
(sin of given angle) = 0.25882