# A projectile is shot at a velocity of  35 m/s and an angle of pi/6 . What is the projectile's peak height?

Feb 2, 2016

15.6 m

#### Explanation:

Equation for the peak height is
H =${v}^{2} {\sin}^{2} \frac{\theta}{2 g}$

This is derived using the fundamental equations of motion and the assumption that the trajectory is parabolic. The objects vertical velocity at the peak height is zero m/s.

v = 35 m/s
Theta =30 degrees
g=9.8m/s^2 Gravitational acceleration