# Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises: 3.22

(a) The balloon was at a height of 223 meters. (b) y = 123 m (c) d = 144 m (d) (i) relative to an observer in the basket: $v_x = 15.0~m/s$ $v_y = 49.0~m/s$ (downward) (ii) relative to an observer on the ground: $v_x = 15.0~m/s$ $v_y = 69.0~m/s$ (downward)

#### Work Step by Step

(a) We can find the initial height of the rock $y_0$. $y = y_0+v_{0y}t+\frac{1}{2}gt^2$ $0-y_0 = (-20.0~m/s)(5.00~s)+\frac{1}{2}(-9.80~m/s^2)(5.00~s)^2$ $y_0 = 223~m$ The balloon was at a height of 223 meters. (b) $y = y_0 + vt$ $y = 223~m + (-20.0~m/s)(5.00~s)$ $y = 123~m$ (c) We can find the horizontal distance the rock travels in 5.00 seconds. $x = v_x~t$ $x = (15.0~m/s)(5.00~s)$ $x = 75.0~m$ We can use $x$ and $y$ to find the distance $d$. $d = \sqrt{(x)^2+(y)^2}$ $d = \sqrt{(75.0~m)^2+(123~m)^2}$ $d = 144~m$ (d) (i) relative to an observer in the basket: $v_x = 15.0~m/s$ $v_y = gt$ $v_y = (-9.80~m/s^2)(5.00~s)$ $v_y = -49.0~m/s$ The vertical velocity relative to an observer in the basket is 49.0 m/s (downward). (ii) relative to an observer on the ground: $v_x = 15.0~m/s$ $v_y = v_{0,y}+gt$ $v_y = -20~m/s-(9.80~m/s^2)(5.00~s)$ $v_y = -69.0~m/s$ The vertical velocity relative to an observer on the ground is 69.0 m/s (downward).

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