Question

A projectile is shot at an angle of `pi/4` and a velocity of `47 m/s`. How far away will the projectile land?

Answer 1

The distance is `=225.3m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=47*sin(1/4pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=47sin(1/4pi)-g*t`

`t=47*1/g*sin(1/4pi)`

`=3.39s`

Resolving in the horizontal direction `rarr^+`

We apply the equation of motion

`s=2u_x*t`

`=2*47cos(1/4pi)*3.39`

`=225.3m`