Question

A projectile is shot at an angle of `pi/12` and a velocity of `45 m/s`. How far away will the projectile land?

Answer 1

Range`=103.316m`

Explanation:

The range of a projectile is given by
`R=(v_0^2*Sin2theta)/g`

Where `R` is the range, `v_0` is the initial or muzzle velocity, `theta` is the throwing angle and `g` is the acceleration due to gravity.
Here `v_0=45m/s`, `theta=pi/12`, `g=9.8m/s^2` and `R=??`

`implies R=(45^2*Sin(2*pi/12))/9.8=(2025*Sin(pi/6))/9.8=((2025)0.5)/9.8=103.316 m`
i
`impliesR=103.316m`