# A ball with a mass of 420 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 18 (kg)/s^2 and was compressed by 6/5 m when the ball was released. How high will the ball go?

Apr 16, 2016

$\approx 2.62 m$

#### Explanation:

By conservation of mechanical energy
PE of the ball at its maximum height when projected by the spring will be equal to PE of the contracted spring
If h m be the max height attained by the ball then

$m g h = \frac{1}{2} \times k \times {x}^{2}$
where

• m= mass = 0.42kg
• g = acceleration due to gravity $= 9.8 \frac{m}{s} ^ 2$
• h = maximum height =?
• k = force constant of spring = $18 \frac{k g}{s} ^ 2$
• x=compression of spring =$\frac{6}{5} m = 1.2 m$

$h = \frac{k {x}^{2}}{2 m g} = \frac{18 \times 1.2}{2 \times 0.42 \times 9.8} \approx 2.62 m$