## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Note that upward is the positive direction. (a) $v^2 = v_0^2+2ay$ $v = \sqrt{v_0^2+2ay}$ $v = \sqrt{(20~m/s)^2+(2)(-9.80~m/s^2)(-10~m)}$ $v = -24.4~m/s$ The rock's velocity is -24.4 m/s (b) $v=v_0+at$ $t = \frac{v-v_0}{a}$ $t = \frac{(-24.4~m/s)-(20~m/s)}{-9.80~m/s^2}$ $t = 4.5~s$ The rock is in the air for 4.5 seconds.