Well, first you must realize that the speed whilst leaving theramp is going to be the same as the start because there is nochange in height. Knowing that, it will be the same type ofproblem as someone jumping in the air. So...

\(\displaystyle{2}={{v}_{{0}}^{{2}}}+{2}{a}{d}\)

(where v=final velocity, \(\displaystyle{v}_{{0}}=\) starting velocity, a=acceleration,d= displacement)

\(\displaystyle{0}\frac{{m}}{{s}}={\left({5.4}\frac{{m}}{{s}^{{2}}}\right)}+{2}{\left(-{9.8}\frac{{m}}{{s}^{{2}}}\right)}{d}-{29.16}\frac{{m}}{{s}^{{2}}}=-{19.6}\frac{{m}}{{s}^{{2}}}\)

d) 1.487755... m = d [using significantfigures] d=1.5 m

\(\displaystyle{2}={{v}_{{0}}^{{2}}}+{2}{a}{d}\)

(where v=final velocity, \(\displaystyle{v}_{{0}}=\) starting velocity, a=acceleration,d= displacement)

\(\displaystyle{0}\frac{{m}}{{s}}={\left({5.4}\frac{{m}}{{s}^{{2}}}\right)}+{2}{\left(-{9.8}\frac{{m}}{{s}^{{2}}}\right)}{d}-{29.16}\frac{{m}}{{s}^{{2}}}=-{19.6}\frac{{m}}{{s}^{{2}}}\)

d) 1.487755... m = d [using significantfigures] d=1.5 m