Question

A projectile is shot at an angle of `pi/4` and a velocity of `8 m/s`. How far away will the projectile land?

Answer 1

`R=6.53m`

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=8m/s`, `theta=pi/4`, `g=9.8m/s^2` and `R=??`

`implies R=(8^2*Sin(2*pi/4))/9.8=(64*Sin(pi/2))/9.8=(64*1)/9.8=6.53 m`
i
`impliesR=6.53m`