A ball with a mass of $480 g$ $480 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $21 \frac{k g}{s^{2}}$ $21 (kg)/s^2$ and was compressed by $\frac{6}{5} m$ $6/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-04-07T07:39:11

The height is $= 3.21 m$ $=3.21m$

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The spring constant is $k = 21 k g s^{- 2}$ $k=21kgs^-2$

The compression is $x = \frac{6}{5} m$ $x=6/5m$

The potential energy is

$P E = \frac{1}{2} \cdot 21 \cdot {(\frac{6}{5})}^{2} = 15.12 J$ $PE=1/2*21*(6/5)^2=15.12J$

This potential energy will be converted to kinetic energy when the spring is released

$K E = \frac{1}{2} m u^{2}$ $KE=1/2m u^2$

The initial velocity is $= u$ $=u$

$u^{2} = \frac{2}{m} \cdot K E = \frac{2}{m} \cdot P E$ $u^2=2/m*KE=2/m*PE$

$u^{2} = \frac{2}{0.48} \cdot 15.12 = 63$ $u^2=2/0.48*15.12=63$

$u = \sqrt{63} = 7.94 m s^{- 1}$ $u=sqrt63=7.94ms^-1$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

We apply the equation of motion

$v^{2} = u^{2} + 2 a h$ $v^2=u^2+2ah$

At the greatest height, $v = 0$ $v=0$

and $a = - g$ $a=-g$

So,

$0 = 63 - 2 \cdot 9.8 \cdot h$ $0=63-2*9.8*h$

$h = \frac{63}{2 \cdot 9.8} = 3.21 m$ $h=63/(2*9.8)=3.21m$

A ball with a mass of 480g480 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 21kgs221 (kg)/s^2 and was compressed by 65m6/5 m when the ball was released. How high will the ball go?