A ball with a mass of $144 g$ $144 g$ 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?

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Answer 1 · 2018-01-01T18:47:36

$h = 2.04 m$ $h=2.04m" "$ (2dps)

when the ball is released from the spring all the Elastic Potential energy will be given to Gravitational Potential energy at maximum height.

$E P E = \frac{1}{2} k x^{2} = \frac{1}{2} \times 16 \times {(\frac{3}{5})}^{2}$ $EPE=1/2kx^2=1/2xx16xx(3/5)^2$

$E P E = \frac{72}{25} J$ $EPE=72/25J$

$now G P E = E P E$ $"now "GPE=EPE$

$m g h = \frac{72}{25}$ $mgh=72/25$

$0.144 \times 9.8 \times h = \frac{72}{25}$ $0.144xx9.8xxh=72/25$

$1.4112 \times h = \frac{72}{25}$ $1.4112xxh=72/25$

$:.h=72/(25xx1.4112)$

$h=2.04m" "$ (2dps)

A ball with a mass of 144 g144g144 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?