# A ball with a mass of 400 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 25 (kg)/s^2 and was compressed by 3/7 m when the ball was released. How high will the ball go?

Sep 26, 2017

Potential Energy stored in spring if it compressed by x meter
$U = \frac{1}{2} k {x}^{2}$
where k$= \text{spring constant}$
$x = \text{Compression}$
$m = 0.4 k g$
$U = \frac{1}{2} \left(25\right) {\left(\frac{3}{7}\right)}^{2} = \frac{25 \times 9}{2 \times 49}$

if spring release the ball will get kinetic energy which is equal to the potential energy of spring. (Conservation of energy)
$K = \frac{1}{2} m {v}^{2} = \frac{25 \times 9}{2 \times 49}$
$\implies 0.4 {v}^{2} = \frac{25 \times 9}{49}$
$\implies {v}^{2} = \frac{25 \times 9}{0.4 \times 49} = \frac{25 \times 9 \times 10}{4 \times 49}$
=>v=sqrt((25xx9xx10)/(4xx49)
$\implies v = \frac{5 \times 3 \times \sqrt{10}}{2 \times 7} = \frac{15}{14} \sqrt{10}$

To find the height we use Newton's Equation
${v}^{2} = {u}^{2} + 2 a s$
where $u =$initial velocity $= \frac{15}{14} \sqrt{10}$
$v = \text{final velocity} = 0$
$a = - g = \text{acceleration due to gravity} = - 9.8 \frac{m}{s} ^ 2$

Hence ${0}^{2} = {\left(\frac{15}{14}\right)}^{2} \times 10 - 2 \left(9.8\right) s$
$s = {\left(\frac{15}{14}\right)}^{2} \left(\frac{10}{2 \times 9.8}\right) = \frac{2250}{3841.6} = 0.585 m$