A projectile is shot at a velocity of  1 m/s and an angle of pi/6 . What is the projectile's peak height?

Nov 6, 2017

H=1/8g m

Explanation:

Let us consider only the vertical motion of the projectile .Let it's velocity be v then it's velocity along the vertical direction is vsin $\theta$

where $\theta$ is the angle made by the velocity of the projectile along the horizontal direction .

The only acceleration along the vertical direction is the acceleration due to gravity g.

At the peak point the velocity of the projectile becomes zero . therefore the final velocity of the projectile is zero .

Let us use the kinematic equation ${v}^{2} = {u}^{2} - 2 g H$

where u is the initial velocity v is the final velocity and H is the maximum height reached by the projectile .

$0 = {\left(u \sin \theta\right)}^{2} - 2 g H$

$H = \frac{{\left(u \sin \left(\frac{\pi}{6}\right)\right)}^{2}}{g \cdot 2}$
$H = \frac{1}{8 g} m$ is the maximum height reached by the projectile