# A ball with a mass of 280 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16 (kg)/s^2 and was compressed by 3/5 m when the ball was released. How high will the ball go?

May 26, 2016

$h = {v}^{2} / \left(2 g\right) = {\left(4.53557\right)}^{2} / 20 = 1.02 \left(\approx\right)$

#### Explanation:

$K {E}_{\text{Ball after release") = PE _("Spring}}$

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2}$

$m {v}^{2} = k {x}^{2}$

$\left(0.28\right) {v}^{2} = 16 \cdot {\left(\frac{3}{5}\right)}^{2}$

Solving

$v = 4.53557 \frac{m}{s}$
$K {E}_{\text{INITIAL") =PE_("FINAL}}$

$\frac{1}{2} \cancel{m} {v}^{2} = \cancel{m} g h$

$h = {v}^{2} / \left(2 g\right) = {\left(4.53557\right)}^{2} / 20 = 1.02 \left(\approx\right)$