# A ball with a mass of 144 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 6/4 m when the ball was released. How high will the ball go?

Apr 23, 2017

The height is $= 12.76 m$

#### Explanation: The spring constant is $k = 16 k g {s}^{-} 2$

The compression is $x = \frac{6}{4} m$

The potential energy is

$P E = \frac{1}{2} \cdot 16 \cdot {\left(\frac{6}{4}\right)}^{2} = 18 J$

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 $= u$

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

${u}^{2} = \frac{2}{0.144} \cdot 18 = 250$

$u = \sqrt{250} = 15.81 m {s}^{-} 1$

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,

$0 = 250 - 2 \cdot 9.8 \cdot h$

$h = \frac{250}{2 \cdot 9.8} = 12.76 m$