Question

A projectile is shot at a velocity of `4 m/s` and an angle of `pi/12`. What is the projectile's maximum height?

Answer 1

`h=((4 times 0.25882)^2)/(2times 9.81) m`
`approx 0.05463 m`

Explanation:

Maximum height is reached because of `Sin theta` component of the velocity.
Use the formula
`v^2-u^2=2gh`
Given `v=0, u= 4 . sin (pi / 12)`; as `theta=pi/12`
Take `g= 9.81 m/s^2`
Since gravity is acting against the velocity so it is deceleration and `-` sign is used in front of g
Substituting various values
`h=((4 . sin (pi / 12))^2)/(2times 9.81) m`
Use tables to substitute the value of
(sin of given angle) = 0.25882