Question

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

Answer 1

15.6 m

Explanation:

See this URL https://en.m.wikipedia.org/wiki/Projectile_motion

Equation for the peak height is
H =`v^2sin^2 (theta)/(2g)`

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