Explanation:
split the situation into two parts
part one - from initial starting point to top of parabola
part two - top of parabola to impact on ground
* v(0) is initial speed
* t(1) is time for part one (from initial start to top)
* y is distance from starting point to the top
* t(2) is time for part 2 (from top to impact)
[tex]v(0) = v(top) + gt(1)[/tex]
[tex]t(1) = \frac{v(0)}{g }[/tex]
[tex]y = v(0)t(1) + \frac{1}{2} g {t(1)}^{2} [/tex]
[tex]y = \ \frac{ {v(0)}^{2} }{g} + \frac{1}{2} g \frac{ {v(0)}^{2} }{ {g}^{2} } [/tex]
[tex]y = \frac{2 {v(0)}^{2} }{3g} [/tex]
[tex]d = y + 1.5[/tex]
[tex]d = \frac{2 {v(0)}^{2} }{3g} + 1.5[/tex]
[tex]d = v(top)t(2) + \frac{1}{2} g {t(2)}^{2} [/tex]
[tex]t(2) = \sqrt{2 (\frac{2 {v(0)}^{2} }{3g} + 1.5) \div g} [/tex]
[tex]t(total) = t(1) + t(2)[/tex]
[tex]t(total) = v(0) + \sqrt{ (\frac{4 {v(0)}^{2} }{3g} + 3) \div g } [/tex]
t(total) =.88 (if you rounded earlier in the problem you would get .87
