A projectile is shot at a velocity of 16 m/s and an angle of pi/6 . What is the projectile's peak height?

1 Answer
Dec 27, 2017

The peak height is =3.27m

Explanation:

Resolving in the vertical direction uarr^+

The initial velocity is u=16sin(1/6pi)ms^-1

Let the peak height be =hm

At the greatest heignt v=0ms^-1

The acceleration due to gravity is g=9.8ms^-2

Applying the equation of motion

v^2=u^2+2as=u^2-2gh

h=(u^2-v^2)/(2g)=((16sin(1/6pi))^2-0)/(2g)=3.27m