[tex]t = 5.7s \\ v0 = 0 \frac{m}{s} \\ a = - 9.8 \frac{m}{ {s}^{2} } [/tex] we want to find x!
the only kineamtic that works with this info is the third! [tex]x = v0 \times t + \frac{1}{2} \times a \times {t}^{2} [/tex] so we plug in our values and get [tex]x = 0 + \frac{1}{2} \times ( - 9.8 \frac{m}{ {s}^{2} } ) \times ( {(5.7s)}^{2} )[/tex] this'll equal out to -159.2 meters.
our units equal out, so we know our answer is correct!
