# A ball with a mass of 160 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 24 (kg)/s^2 and was compressed by 1/4 m when the ball was released. How high will the ball go?

Apr 23, 2017

The height of the ball is $= 0.48 m$

#### Explanation: The spring constant is $k = 24 k g {s}^{-} 2$

The compression is $x = \frac{1}{4} m$

The potential energy is

$P E = \frac{1}{2} \cdot 24 \cdot {\left(\frac{1}{4}\right)}^{2} = \frac{3}{4} J$

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 $= u$

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

${u}^{2} = \frac{2}{0.16} \cdot \frac{3}{4} = 9.375$

$u = \sqrt{9.375} = 3.06 m {s}^{-} 1$

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,

$0 = 9.375 - 2 \cdot 9.8 \cdot h$

$h = \frac{9.375}{2 \cdot 9.8} = 0.48 m$
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