Question

A projectile is shot at an angle of `pi/3` and a velocity of `28 m/s`. How far away will the projectile land?

Answer 1

`69.3m`

Explanation:

I'll present a "shortcut" equation to find the distance a projectile will launch given an initial launch velocity and angle. The formula for maximum range `R` of a launched projectile is given by the equation

`R = ((v_0)^2 sin(2alpha_0))/g`

where `v_0` is the magnitude of the initial velocity,
`alpha_0` is the initial launch angle, and
`g` is the acceleration to to gravity near Earth's surface, `9.8m/(s^2)`.

Although it doesn't necessarily matter, the equivalent angle of `pi/3` is `60^o`. We can plug in the known variables to solve for the maximum range `R`:

`R = ((28m/s)^2 sin(2(60^o)))/(9.8m/(s^2)) = color(blue)(69.3m`