A ball with a mass of $450 g$ $450 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $15 \frac{k g}{s^{2}}$ $15 (kg)/s^2$ and was compressed by $\frac{8}{3} m$ $8/3 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-08-07T07:28:58

The height is $= 12.09 m$ $=12.09m$

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The spring constant is $k = 15 k g s^{- 2}$ $k=15kgs^-2$

The compression is $x = \frac{8}{3} m$ $x=8/3m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 15 \cdot {(\frac{8}{3})}^{2} = 53.33 J$ $PE=1/2*15*(8/3)^2=53.33J$

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

$K E_{b a l l} = \frac{1}{2} m u^{2}$ $KE_(ball)=1/2m u^2$

Let the height of the ball be $= h$ $=h$

Then ,

The potential energy of the ball is $P E_{b a l l} = m g h$ $PE_(ball)=mgh$

$P E_{b a l l} = 53.33 = 0.450 \cdot 9.8 \cdot h$ $PE_(ball)=53.33=0.450*9.8*h$

$h = 53.33 \cdot \frac{1}{0.450 \cdot 9.8}$ $h=53.33*1/(0.450*9.8)$

$= 12.09 m$ $=12.09m$

A ball with a mass of 450g450 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 15kgs215 (kg)/s^2 and was compressed by 83m8/3 m when the ball was released. How high will the ball go?