Question

A projectile is shot at an angle of `pi/12` and a velocity of `9 m/s`. How far away will the projectile land?

Answer 1

Distance to a landing spot is
`V_h*2*V_v/g ~~ 4.1326 (m)`
where
`V_h = 9m/(sec)*cos(pi/12)`
`V_v = 9m/(sec)*sin(pi/12)`

Explanation:

Vector of velocity of a projectile should be represented as a sum of two vectors - horizontal and vertical components.
These components are:
`V_h = 9m/(sec)*cos(pi/12) ~~ 8.69333m/sec`
`V_v = 9m/(sec)*sin(pi/12) ~~2.32937 m/sec`

While the vertical component brings the projectile up, after which it falls down, the horizontal component pulls it forward during the same time.

So, to find the distance to a landing spot, we need the time the projectile is flying.
On its way up the projectile's vertical component of the velocity is diminishing every second by `g m/(sec^2) ~~ 9.8 m/sec^2`.
Therefore, with initial vertical component of `V_v = 9m/(sec)*sin(pi/12)` and a loss of `9.8 m/sec^2` every second, the projectile will continue rising for
`V_v/g ~~ (9sin(pi/12))/9.8 ~~ 0.23769 (sec)`

The same time is needed by the projectile to come back to earth, which brings the total time to go up and down to
`t = 0.23769 * 2 = 0.47538 (sec)`

During the same vertical component was dragging the projectile forward with a speed `V_h`.

So, the distance covered is
`V_h*0.47538=9cos(pi/12)*0.47538 ~~4.1326 m`