$\Delta y = v_0 \sin{\theta} \cdot t - \dfrac{1}{2}gt^2$
Using $g = 9.8 \, m/s^2$ ...
$-45=v_0 \sin(-30) \cdot t - 4.9 t^2$
$0=45 + v_0 \cdot \left(-\dfrac{1}{2}\right) \cdot t - 4.9 t^2$
$0=90-v_0 \cdot t - 9.8t^2$
$\Delta x = v_0\cos{\theta} \cdot t$
$57=v_0 \cos(-30) \cdot t$
$57=v_0 \cdot \dfrac{\sqrt{3}}{2} \cdot t \implies v_0 \cdot t = \dfrac{114}{\sqrt{3}}$
substitute $\dfrac{114}{\sqrt{3}}$ for $v_0 \cdot t$ in the $\Delta y$ equation and solve for $t$.
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Last edited by skeeter; Apr 1st 2017 at 10:42 AM.
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