A ball with a mass of $450 g$ $450 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $12 \frac{k g}{s^{2}}$ $12 (kg)/s^2$ and was compressed by $\frac{4}{5} m$ $4/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2016-07-05T09:23:06

$h = \frac{128}{15 g}$ $h = 128/(15 g)$ [m]

Calling

$h$ $h$ height attained
$m = 0.450$ $m = 0.450$ ball mass
$K = 12$ $K=12$ spring constant
$Delta x = 4/5$ compressed length

Equating potential energies

$m g h = 1/2K (Delta x)^2$

then

$h = 1/2 K/(mg) (Deltax)^2 = 128/(15 g)$

A ball with a mass of 450 g450g450 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 12 (kg)/s^212kgs212 (kg)/s^2 and was compressed by 4/5 m45m4/5 m when the ball was released. How high will the ball go?