Question

If a projectile is shot at a velocity of `3 m/s` and an angle of `pi/3`, how far will the projectile travel before landing?

Answer 1

Approximately `0.8` meters.

Explanation:

I assume that the projectile was launched from flat ground.

Range of a projectile is given by the equation:

`d=(v^2sin(2theta))/g`

where:

• `d` is the total distance traveled

• `v` is the initial velocity

• `theta` is the angle of incline

• `g` is the gravitational acceleration, which is `9.8 \ "m/s"^2` on Earth

So, we get:

`d=((3 \ "m/s")^2*sin((2pi)/3))/(9.8 \ "m/s"^2)`

`=(9 \ "m"^2"/s"^2*0.87)/(9.8 \ "m/s"^2)`

`=(7.83color(red)cancelcolor(black)"m"^2"/"color(red)cancelcolor(black)("s"^2))/(9.8color(red)cancelcolor(black)"m""/"color(red)cancelcolor(black)("s"^2))`

`~~0.8 \ "m"`