# 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 42 (kg)/s^2 and was compressed by 5/3 m when the ball was released. How high will the ball go?

Apr 12, 2016

#### Answer:

$h = 21 , 25 \text{ m}$

#### Explanation:

$\text{given data:}$

$\Delta x = \frac{5}{3} m$

$K = 42 \frac{k g}{s} ^ 2$

$m = 280 g = 0 , 28 k g \text{ mass of object}$

$g = 9 , 81 \frac{m}{s} ^ 2$

h=?

${E}_{p} = \frac{1}{2} \cdot k \cdot \Delta {x}^{2} \text{ Potential energy for compressed springs}$

$E = m \cdot g \cdot h \text{ potential energy for object raising from earth}$

${E}_{p} = E \text{ conservation of energy}$

$m \cdot g \cdot h = \frac{1}{2} \cdot K \cdot \Delta {x}^{2}$

$h = \frac{K \cdot \Delta {x}^{2}}{2 \cdot m \cdot g}$

$h = \frac{42 \cdot {\left(\frac{5}{3}\right)}^{2}}{2 \cdot 0 , 28 \cdot 9 , 81}$

$h = \frac{42 \cdot \frac{25}{9}}{5 , 49}$

$h = \frac{116 , 67}{5 , 49}$

$h = 21 , 25 \text{ m}$