# 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 16 (kg)/s^2 and was compressed by 3/7 m when the ball was released. How high will the ball go?

Jun 17, 2016

$h = \frac{E p}{m g} = \frac{E s}{m g} = \frac{0.5 k {x}^{2}}{m g} = 0.375 m$

#### Explanation:

The ball is on a compressed spring; it has stored potential energy from the spring. This turns into kinetic energy and launches the ball in the air. Finally, once the ball is at it's highest point it will temporarily be stationary; it will have 0 kinetic energy and therefore all the energy will be gravitational potential energy!
(We are ignoring the effects of air resistance here).

Potential energy in a spring;
$E s = 0.5 k {x}^{2}$
where k is spring constant, and x is compression distance.

We can ignore the kinetic energy as this will be converted to gravitational potential energy.

Gravitational potential energy;
$E p = m g h$
where m is mass of ball, g is assumed 9.81$m {s}^{-} 2$, and h is height.

We can rearrange to solve for h given that Ep = Es.
$h = \frac{E p}{m g} = \frac{E s}{m g} = \frac{0.5 k {x}^{2}}{m g} = \frac{0.5 \cdot 16 \cdot {\left(\frac{3}{7}\right)}^{2}}{0.4 \cdot 9.81} = 0.375 m$