Question

A projectile is shot at an angle of `pi/4` and a velocity of `19 m/s`. How far away will the projectile land?

Answer 1

The range is `=36.1 m`

Explanation:

We use two formulas
`x=v_0costheta*t`
`h=v_0sintheta*t-1/2g*t^2`
Here we have `v_0=19m/s` and `theta=pi/4`
`h=0`
`0=v_0sintheta*t-1/2g*t^2`
`:.t=0` or `t=2v_0sintheta/g`
So `x=v_0^2*2sintheta*costheta/g=(v_0^2*sin2theta)/g`
`x=19*19*sin(pi/2)*1/g=361/10=36.1 m`