## College Physics (4th Edition)

The penny hits the ground with a velocity of $85.0~m/s$
We can find the velocity $v_f$ when the penny hits the ground: $v_f^2 = v_0^2+2a\Delta y$ $v_f = \sqrt{v_0^2+2a\Delta y}$ $v_f = \sqrt{0+(2)(9.80~m/s^2)(369~m)}$ $v_f = 85.0~m/s$ The penny hits the ground with a velocity of $85.0~m/s$