# A ball with a mass of 600 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 3/5 m when the ball was released. How high will the ball go?

Apr 2, 2017

The height is $= 0.49 m$

#### Explanation:

The energy stored in the spring will be converted to kinetic energy and this will give the initial speed of the ball.

Mass is $m = 0.6 k g$

Spring constant is $k = 16 k g {s}^{-} 2$

Compression is $x = \frac{3}{5} m$

Energy in the spring ${E}_{s} = \frac{1}{2} k {x}^{2}$

Kinetic energy is $K E = \frac{1}{2} m {v}^{2}$

$\frac{1}{2} k {x}^{2} = \frac{1}{2} m {v}^{2}$

${v}^{2} = \frac{k}{m} {x}^{2}$

$= \frac{16}{0.6} \cdot {\left(\frac{3}{5}\right)}^{2}$

$= 9.6 {m}^{2} {s}^{-} 2$

$v = \sqrt{9.6} = 3.1 m {s}^{-} 1$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

To find the greatest height

$v = 0$

$u = 3.1$

$a = - g = - 9.8$

$s = h$

So,

$0 = {3.1}^{2} - 2 \cdot 9.8 \cdot h$

$h = \frac{9.6}{2 \cdot 9.8}$

$h = 0.49 m$