#### Answer

(a) The speed is 6.69 m/s.
(b) It goes 4.48 meters above the ground.
(c) The time it takes to fall from the maximum height down to the height of the athlete's head is 0.735 seconds.

#### Work Step by Step

(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.