# Chapter 4 - Problems - Page 149: 34

(a) The stone rises to a maximum height of $21.1~m$ above the ground. (b) The total time that elapses until the stone hits the ground is $4.08~s$

#### Work Step by Step

(a) At maximum height, we can find the displacement $\Delta y$ above the initial vertical position of $1.50~m$: $v_f^2 = v_0^2+2a\Delta y$ $\Delta y = \frac{v_f^2-v_0^2}{2a}$ $\Delta y = \frac{0-(19.6~m/s)^2}{(2)(-9.80~m/s^2)}$ $\Delta y = 19.6~m$ The stone rises to a maximum height of $19.6~m + 1.50~m$ which is $21.1~m$ above the ground. (b) We can find the time to reach maximum height: $v_f = v_0+at$ $t = \frac{v_f-v_0}{a}$ $t = \frac{0-19.6~m/s}{-9.80~m/s^2}$ $t = 2.0~s$ We can find the time it takes to drop from the maximum height to the ground: $\Delta y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2\Delta y}{a}}$ $t = \sqrt{\frac{(2)(-21.1~m)}{-9.80~m/s^2}}$ $t = 2.08~s$ The total time that elapses until the stone hits the ground is $2.0~s+2.08~s$ which is $4.08~s$

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