#### Answer

(a) $F = 225~N$
(b) $v = 0.200~m/s$

#### Work Step by Step

(a) We can find the acceleration of the rock.
$a = \frac{v^2-v_0^2}{2d}$
$a = \frac{(30~m/s)^2-0}{(2)(1.0~m)}$
$a = 450~m/s^2$
We can use the acceleration to find the force that Bob exerts on the rock.
$F = ma$
$F = (0.500~kg)(450~m/s^2)$
$F = 225~N$
(b) By Newton's third law, the rock exerts an equal and opposite force on Bob. We can find Bob's acceleration.
$F = ma$
$a = \frac{F}{m}$
$a = \frac{225~N}{75~kg}$
$a = 3.0~m/s^2$
We can find the time it takes for the rock to move through a distance of 1.0 meter while Bob is throwing it.
$d = \frac{1}{2}at^2$
$t = \sqrt{\frac{2d}{a}}$
$t = \sqrt{\frac{(2)(1.0~m)}{450~m/s^2}}$
$t = 0.0667~s$
We can find Bob's speed after releasing the rock.
$v = v_0+at$
$v = 0 + (3.0~m/s^2)(0.0667~s)$
$v = 0.200~m/s$