# A ball with a mass of 70 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 5 (kg)/s^2 and was compressed by 7/6 m when the ball was released. How high will the ball go?

Apr 6, 2018

#### Answer:

The height is $= 4.96 m$

#### Explanation:

The spring constant is $k = 5 k g {s}^{-} 2$

The compression is $x = \frac{7}{6} m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 5 \cdot {\left(\frac{7}{6}\right)}^{2} = 3.40 J$

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

$K {E}_{b a l l} = \frac{1}{2} m {u}^{2}$

Let the height of the ball be $= h$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Then ,

The potential energy of the ball is $P {E}_{b a l l} = m g h$

Mass of the ball is $m = 0.070 k g$

$P {E}_{b a l l} = 3.40 = 0.070 \cdot 9.8 \cdot h$

$h = 3.40 \cdot \frac{1}{0.07 \cdot 9.8}$

$= 4.96 m$

The height is $= 4.96 m$