Answer
33.3 m/s
Work Step by Step
Since the mechanical energy $E_{mec}$ is conserved, we have that $\Delta E_{mec} = \Delta K + \Delta U = \frac{1}{2}m(v^{2}–v_{0}^{2}) + mg\Delta h$ = 0.
From the given info, we know that m = 825 kg, $v_{0}$ = 17.0 m/s and h = 42.0 m. Looking at the figure, we see that at point C the car has dropped a height h, and therefore the change in height $\Delta h$ = – h = – 42.0 m. So solving for v, the desired speed, and plugging in our values, we get:
$\frac{1}{2}m(v^{2}–v_{0}^{2}) + mg\Delta h$ = 0
$\frac{1}{2}m(v^{2}–v_{0}^{2}) = –mg\Delta h$
$v^{2}–v_{0}^{2}$ = $–2g\Delta h$
$v^{2}$ = $–2g\Delta h + v_{0}^{2}$
$v= \sqrt {–2g\Delta h + v_{0}^{2}}$
= $\sqrt {–2(9.8 m/s^{2})(–42.0m) + (17.0 m/s)^{2}}$
= $33.3$ $ m/s$
