# A ball with a mass of 640 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32 (kg)/s^2 and was compressed by 5/8 m when the ball was released. How high will the ball go?

Jan 12, 2018

0.99 meters

#### Explanation:

When the spring was released after compression,the elastic potential energy of the spring got transferred to the ball as it's kinetic energy and the spring came to its original position.
So,we can say
$\left(\frac{1}{2}\right)$K$\left({x}^{2}\right)$=$\left(\frac{1}{2}\right)$m$\left({u}^{2}\right)$
(k is the spring constant,x is the length of compression,m is mass of the ball and u is its velocity at the beginning of the journey upwards)
Putting the values we get,$\left({u}^{2}\right)$=19.53 m/s
So,using equation of velocity,
${v}^{2} = {u}^{2} - 2 a s$(all symbols are bearing their conventional meaning)
Here,v=0,${u}^{2}$=19.53,a=g,s is the desired height
We get,s=0.99 meters

Alternatively,
We can say after reaching upto height s,this amount of kinetic energy attributed to the ball due to compression of the spring will get converted to its potential energy.
Hence,
$\left(\frac{1}{2}\right)$k$\left({x}^{2}\right)$=mgs
So,s=0.99 meters