Question

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

Answer 1

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^2-2gH`

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

`0=(usintheta)^2-2gH`

`H=((usin(pi/6))^2)/(g*2)`
`H=1/(8g) m` is the maximum height reached by the projectile