# A ball with a mass of 80 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 1/5 m when the ball was released. How high will the ball go?

Jan 8, 2017

I tried this but check my maths.

#### Explanation:

We can consider:

So basically we can write:
$\frac{1}{2} k {x}^{2} = \frac{1}{2} m {v}^{2}$
$v = \sqrt{\frac{k {x}^{2}}{m}} = \sqrt{8} = 2.8 \frac{m}{s}$

We now use the kinematic relationship:
${v}_{f}^{2} = {v}_{i}^{2} + 2 a h$
With $a =$ acceleration of gravity downwards and ${v}_{f} = 0$ for the final velocity;
$0 = 8 - \left(2 \cdot 9.8 \cdot h\right)$
$h = \frac{8}{2 \cdot 9.8} = 0.4 m$
If we consider also the initial compression, the total vertical distance should be:
$h + {h}_{1} = 0.4 + 0.2 = 0.6 m$