## University Physics with Modern Physics (14th Edition)

(a) Let $v_0$ be the speed at the point of release. Let $y$ be the maximum height. $y = \frac{v_0^2}{2g}$ Let $v$ be the speed when the object returns to the point of release. $v^2 = 0 + 2gy = 2g(\frac{v_0^2}{2g}) = v_0^2$ $v = v_0$ The object will have the same speed when it returns to the point of release. (b) Let $t_1$ be the time for the object to rise to its maximum height $y$. $t_1 = \frac{v_0}{g}$ Let $t_2$ be the time for the object to fall back down to the point of release. $t_2 = \frac{v}{g}$ Let $t$ be the total time of flight. $t = t_1+t_2 = \frac{v_0}{g} + \frac{v}{g}$ $t = \frac{v_0}{g}+\frac{v_0}{g} = 2\times \frac{v_0}{g}$ $t = 2\times t_1$ The time of flight will be twice the time to get to the highest point.