## College Physics (4th Edition)

(a) We can find the distance the ball falls in the first $3.0~s$: $\Delta y = \frac{1}{2}at^2$ $\Delta y = \frac{1}{2}(9.80~m/s^2)(3.0~s)^2$ $\Delta y = 44.1~m$ The ball falls 44.1 meters in the first 3.0 seconds. (b) We can find the speed after it has traveled 2.5 m downward: $v_f^2=v_0^2+2a\Delta y$ $v_f= \sqrt{v_0^2+2a\Delta y}$ $v_f= \sqrt{0+(2)(9.80~m/s^2)(2.5~m)}$ $v_f = 7.0~m/s$ The speed of the ball is 7.0 m/s (c) We can find the speed of the ball 3.0 seconds after it is released: $v_f = v_0+at$ $v_f = 0 + (9.80~m/s^2)(3.0~s)$ $v_f = 29.4~m/s$ The speed of the ball 3.0 seconds after it is released is 29.4 m/s