# A ball with a mass of 250 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 8 (kg)/s^2 and was compressed by 3/2 m when the ball was released. How high will the ball go?

May 31, 2018

The height reached by the ball is $= 3.67 m$

#### Explanation: The spring constant is $k = 8 k g {s}^{-} 2$

The compression of the spring is $x = \frac{3}{2} m$

The potential energy stored in the spring is

$P E = \frac{1}{2} k {x}^{2} = \frac{1}{2} \cdot 8 \cdot {\left(\frac{3}{2}\right)}^{2} = 9 J$

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 $m = 0.250 k g$

$P {E}_{b a l l} = 9 = 0.250 \cdot 9.8 \cdot h$

$h = 9 \cdot \frac{1}{0.250 \cdot 9.8}$

$= 3.67 m$

The height reached by the ball is $= 3.67 m$