# A ball with a mass of 125 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 8 (kg)/s^2 and was compressed by 2/5 m when the ball was released. How high will the ball go?

Apr 8, 2017

The height is $= 0.52 m$

#### Explanation:

The spring constant is $k = 8 k g {s}^{-} 2$

The compression is $x = \frac{2}{5} m$

The potential energy is

$P E = \frac{1}{2} \cdot 8 \cdot {\left(\frac{2}{5}\right)}^{2} = \frac{16}{25} = 0.64 J$

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 $= u$

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

${u}^{2} = \frac{2}{0.125} \cdot 0.64 = 10.24$

$u = \sqrt{10.24} = 3.2 m {s}^{-} 1$

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,

$0 = 10.24 - 2 \cdot 9.8 \cdot h$

$h = \frac{10.24}{2 \cdot 9.8} = 0.52 m$