# A ball with a mass of 450 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 4/3 m when the ball was released. How high will the ball go?

May 8, 2017

The height is $= 2.42 m$

#### Explanation: The spring constant is $k = 12 k g {s}^{-} 2$

The compression is $x = \frac{4}{3} m$

The potential energy is

$P E = \frac{1}{2} \cdot 12 \cdot {\left(\frac{4}{3}\right)}^{2} = \frac{32}{3} J$

This potential energy will be converted to kinetic energy when the spring is released

$K E = \frac{1}{2} m {u}^{2}$

The initial velocity is $= u$

${u}^{2} = \frac{2}{m} \cdot K E = \frac{2}{m} \cdot P E$

${u}^{2} = \frac{2}{0.45} \cdot \frac{32}{3} = 47.41$

$u = \sqrt{47.41} = 6.89 m {s}^{-} 1$

Resolving in the vertical direction ${\uparrow}^{+}$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a h$

At the greatest height, $v = 0$

and $a = - g$

So,

$0 = 47.41 - 2 \cdot 9.8 \cdot h$

$h = \frac{47.41}{2 \cdot 9.8} = 2.42 m$