# 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 12 (kg)/s^2 and was compressed by 7/3 m when the ball was released. How high will the ball go?

Jun 20, 2016

$\frac{\text{spring constant "(k)=12"kg}}{s} ^ 2$
$\text{compressed length } \left(x\right) = \frac{7}{3} m$
Potential energy of spring=$\frac{1}{2} k {x}^{2}$

$\text{Mass of the ball } \left(m\right) = 280 g = 0.28 k g$

By principle of conservation of energy

Energy of spring=energy of the ball

$\frac{1}{2} \cdot k \cdot {x}^{2} = m g h$
$\implies h = \frac{1}{2} \frac{k {x}^{2}}{m g}$
$\implies h = \frac{1}{2} \cdot 12 \cdot \left(\frac{7}{3} \cdot \frac{7}{3}\right) \cdot \frac{1}{0.28 \cdot 9.8}$
$\implies h = \frac{1}{3 \cdot 0.028} = 11.9 m$