Answer
(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).
