A ball with a mass of $400 g$ $400 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $25 \frac{k g}{s^{2}}$ $25 (kg)/s^2$ and was compressed by $\frac{3}{7} m$ $3/7 m$ when the ball was released. How high will the ball go?

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A ball with a mass of $400 g$ $400 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $25 \frac{k g}{s^{2}}$ $25 (kg)/s^2$ and was compressed by $\frac{3}{7} m$ $3/7 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-09-26T12:48:23

Potential Energy stored in spring if it compressed by x meter
$U = \frac{1}{2} k x^{2}$ $U=1/2 kx^2$
where k $= spring constant$ $="spring constant"$
$x = Compression$ $x="Compression"$
$m = 0.4 k g$ $m=0.4 kg$
$U = \frac{1}{2} (25) {(\frac{3}{7})}^{2} = \frac{25 \times 9}{2 \times 49}$ $U=1/2(25)(3/7)^2=(25xx9)/(2xx49)$

if spring release the ball will get kinetic energy which is equal to the potential energy of spring. (Conservation of energy)
$K = \frac{1}{2} m v^{2} = \frac{25 \times 9}{2 \times 49}$ $K=1/2mv^2=(25xx9)/(2xx49)$
$\Rightarrow 0.4 v^{2} = \frac{25 \times 9}{49}$ $=>0.4v^2=(25xx9)/(49)$
$\Rightarrow v^{2} = \frac{25 \times 9}{0.4 \times 49} = \frac{25 \times 9 \times 10}{4 \times 49}$ $=>v^2=(25xx9)/(0.4xx49)=(25xx9xx10)/(4xx49)$
$\Rightarrow v = \sqrt{\frac{25 \times 9 \times 10}{4 \times 49}}$ $=>v=sqrt((25xx9xx10)/(4xx49)$
$\Rightarrow v = \frac{5 \times 3 \times \sqrt{10}}{2 \times 7} = \frac{15}{14} \sqrt{10}$ $=>v=(5xx3xxsqrt10)/(2xx7)=15/14sqrt10$

To find the height we use Newton's Equation
$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$
where $u =$ $u=$ initial velocity $= \frac{15}{14} \sqrt{10}$ $=15/14sqrt10$
$v = final velocity = 0$ $v="final velocity"=0$
$a = - g = acceleration due to gravity = - 9.8 \frac{m}{s^{2}}$ $a=-g="acceleration due to gravity"=-9.8m/s^2$

Hence $0^{2} = {(\frac{15}{14})}^{2} \times 10 - 2 (9.8) s$ $0^2=(15/14)^2xx10-2(9.8)s$
$s = {(\frac{15}{14})}^{2} (\frac{10}{2 \times 9.8}) = \frac{2250}{3841.6} = 0.585 m$ $s=(15/14)^2(10/(2xx9.8))=2250/3841.6=0.585 m$

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A ball with a mass of $400 g$ $400 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $25 \frac{k g}{s^{2}}$ $25 (kg)/s^2$ and was compressed by $\frac{3}{7} m$ $3/7 m$ when the ball was released. How high will the ball go?

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A ball with a mass of $400 g$ $400 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $25 \frac{k g}{s^{2}}$ $25 (kg)/s^2$ and was compressed by $\frac{3}{7} m$ $3/7 m$ when the ball was released. How high will the ball go?