Answer
(a) v = 7.47 m/s
(b) The ball flies up to a maximum height of 3.01 meters.
Work Step by Step
(a) We can use conservation of energy to solve this question. Note that when the spring is released and as the ball moves up a distance of 0.160 meters, the initial elastic potential energy is transformed into gravitational potential energy and kinetic energy:
$KE + GPE = EPE$
$\frac{1}{2}mv^2 + mgy = \frac{1}{2}kx^2$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2 - mgy$
$v^2 = \frac{kx^2 - 2mgy}{m}$
$v = \sqrt{\frac{kx^2 - 2mgy}{m}}$
$v = \sqrt{\frac{(875 ~N/m)(0.160 ~m)^2 - (2)(0.380 ~kg)(9.80 ~m/s^2)(0.160 ~m)}{0.380 ~kg}}$
$v = 7.47 ~m/s$
(b) We can use conservation of energy to solve this question. The initial elastic potential energy is transformed into gravitational potential energy at maximum height:
$GPE = EPE$
$mgy = \frac{1}{2}kx^2$
$y = \frac{\frac{1}{2}kx^2}{mg}$
$y = \frac{\frac{1}{2}(875 ~N/m)(0.160 ~m)^2}{(0.380 ~kg)(9.80 ~m/s^2)}$
$y = 3.01~m$
