A ball with a mass of $280 g$ $280 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $16 \frac{k g}{s^{2}}$ $16 (kg)/s^2$ and was compressed by $\frac{3}{5} m$ $3/5 m$ when the ball was released. How high will the ball go?

Question

1740 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-26T09:18:56

$h = \frac{v^{2}}{2 g} = \frac{{(4.53557)}^{2}}{20} = 1.02 (\approx)$ $h = v^2/(2g)= (4.53557)^2/20 =1.02 (approx)$

$K E_{Ball after release} = P E_{Spring}$ $KE_("Ball after release") = PE _("Spring")$

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

$m v^{2} = k x^{2}$ $mv^2 =kx^2$

$(0.28) v^{2} = 16 \cdot {(\frac{3}{5})}^{2}$ $(0.28)v^2 = 16*(3/5)^2$

Solving

$v = 4.53557 \frac{m}{s}$ $v =4.53557 m/s$
$K E_{INITIAL} = P E_{FINAL}$ $KE_("INITIAL") =PE_("FINAL")$

$1/2cancel(m)v^2 =cancelmgh$

$h = v^2/(2g)= (4.53557)^2/20 =1.02 (approx)$

A ball with a mass of 280 g280g280 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16 (kg)/s^216kgs216 (kg)/s^2 and was compressed by 3/5 m35m3/5 m when the ball was released. How high will the ball go?