## Physics: Principles with Applications (7th Edition)

(a) $v = \sqrt{2gH}$ (b) H = 4.8 meters (c) H = 36 meters
(a) $v^2 = v_0^2 + 2gH = 0 + 2gH = 2gH$ $v = \sqrt{2gH}$ (b) $v = (35 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 9.7 ~m/s$ $v = \sqrt{2gH}$ $H = \frac{v^2}{2g} = \frac{(9.7 ~m/s)^2}{(2)(9.80 ~m/s^2)} = 4.8 ~m$ (c) $v = (95 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 26.39 ~m/s$ $v = \sqrt{2gH}$ $H = \frac{v^2}{2g} = \frac{(26.39 ~m/s)^2}{(2)(9.80 ~m/s^2)} = 35.5\approx36 ~m$