## University Physics with Modern Physics (14th Edition)

(a) $v^2 = v_0^2 + 2ay = 0 + 2ay$ $v = \sqrt{2ay} = \sqrt{(2)(35.0~m/s^2)(0.640~m)}$ $v = 6.69~m/s$ The speed is 6.69 m/s. (b) We can let $v_0 = 6.69~m/s$ for this part of the question. $(y-y_0) = \frac{v^2-v_0^2}{2g} = \frac{0-(6.69~m/s)^2}{(2)(-9.80~m/s^2)}$ $y- y_0 = 2.28~m$ $y = 2.28~m + 2.20~m = 4.48~m$ It goes 4.48 meters above the ground. (c) We can find the time it takes for the shot to fall from the maximum height of 4.48 m down to a height of 1.83 m. $y = y_0 - \frac{1}{2}gt^2$ $1.83~m = 4.48~m - (4.9~m/s^2)t^2$ $t^2 = \frac{1.83~m - 4.48~m}{-4.9~m/s^2}$ $t = \sqrt{\frac{1.83~m - 4.48~m}{-4.90~m/s^2}}$ $t = 0.735~s$ The time it takes to fall from the maximum height down to the height of the athlete's head is 0.735 seconds.