Question

A projectile is shot at a velocity of `6 m/s` and an angle of `pi/12`. What is the projectile's maximum height?

Answer 1

`S = (V*sin(phi))^2/(2g)`
where
`V=6m/sec` - initial speed,
`phi=pi/12` - angle to horizon of launching and
`g=9.8m/sec^2` - gravitational constant

The answer (approximately) is `0.123 m`

Explanation:

Initial velocity of `V=6m/sec` directed at an angle `phi=pi/12` to horizon can be represented as a sum two vectors:

`V_v = V*sin(phi)` directed vertically up and
`V_h = V*cos(phi)` directed horizontally.

It's the vertical component `V_v` that drives the projectile up against the constant gravity deceleration `g=9.8m/sec^2`

Since every second the projectile looses its vertical component by `g`, we can calculate the time until its vertical component becomes zero (the top of trajectory):
`t = V_v/g`

Knowing time to the top and initial vertical velocity, the length is calculated by a formula
`S = V_v*t - (g*t^2)/2 = V_v^2/g - (g*V_v^2)/(2g^2) = V_v^2/(2g)`

Incidentally, this is equal to a time `t=V_v/g` multiplied by an average speed, which under constant deceleration from `V_v` to `0` equals to `V_v/2`.

Substituting the real values for `V_v` and `g`,
`S = [6^2*sin^2(pi/12)]/(2*9.8)=[18*(2sin^2(pi/12))]/19.6=`
`=0.92(1-cos(pi/6)) = 0.92*(1-sqrt(3)/2) =`
`0.92*0.134 = 0.123 (m)`