## College Physics (4th Edition)

When the package hits the ground, the horizontal distance between the package and the helicopter is $23~m$
We can find the time to fall $18~m$: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(18~m)}{9.80~m/s^2}}$ $t = 1.917~s$ We can find the horizontal distance between the package and the helicopter when the package hits the ground. Note that we should use the relative velocity between the package and the helicopter for $v_x$: $x = v_x~t = (12~m/s)(1.917~s) = 23~m$ When the package hits the ground, the horizontal distance between the package and the helicopter is $23~m$