# A 15g ball is shot from a spring gun whose spring has force constant of 600N/m. The spring is compressed by 5cm. What is the greatest possible horizontal range of the ball for this compression is (take g=10m/s^2)?

Jan 5, 2018

10 m

#### Explanation:

The energy stored in the spring is given by:

$\textsf{E = \frac{1}{2} k {x}^{2}}$

k is the force constant.

x is the extension.

$\therefore$$\textsf{E = \frac{1}{2} \times 600 \times {0.05}^{2} = 0.75 \textcolor{w h i t e}{x} J}$

I will assume that all this energy will appear as the kinetic energy of the ball:

$\therefore$$\textsf{K E = \frac{1}{2} m {v}^{2}}$

$\textsf{v = \sqrt{\frac{2 K E}{m}} = \sqrt{\frac{2 \times 0.75}{0.015}} = 10 \textcolor{w h i t e}{x} \text{m/s}}$

I will assume that the ball is being launched from the ground as no height is given. The range is given by:

$\textsf{d = \frac{{v}^{2} \sin 2 \theta}{g}}$

Where $\textsf{\theta}$ is the angle of launch. To find the value which will give the maximum range we find the 1st derivative and set it to zero.

Using The Chain Rule:

$\textsf{\frac{d \left(d\right)}{d \left(\theta\right)} = {v}^{2} / g \times 2 \cos 2 \theta = 0}$

$\therefore$$\textsf{\cos 2 \theta = 0}$

$\therefore$$\textsf{2 \theta = \frac{\pi}{2}}$

$\textsf{\theta = \frac{\pi}{4} = {45}^{\circ}}$

This is the launch angle that gives the maximum range.

Using this value gives:

$\textsf{d = \frac{{10}^{2} \times \sin 90}{10} = \frac{100 \times 1}{10} = 10 \textcolor{w h i t e}{x} m}$