# Projectile launcher problem?

## A projectile launcher is set on a table so that the ball becomes a projectile at a height of 1.2 meters above the floor. The mass of the ball is 10 grams (0.01kg). A plunger is used to push the ball into the barrel of the launcher, compressing the spring a distance of 10 cm (.10m) before it is ready to launch. It is launched at an angle of 30 degrees above the horizontal. The ball travels a horizontal distance of 4 meters before striking the floor below. Determine the force constant of the spring that shot the ball in this fashion.

Mar 1, 2018

$29.82$

#### Explanation:

Let ${E}_{k} = \frac{1}{2} m {v}_{0}^{2}$ be the ball kinetic energy leaving the barrel. Then

$\frac{1}{2} K {\left(\Delta x\right)}^{2} = \frac{1}{2} m {v}_{0}^{2}$ so

$K = m {\left({v}_{0} / \left(\Delta x\right)\right)}^{2}$

where

$\Delta x = 0.1$ [m]
$m = 0.01$ [kg]

Follows the ${v}_{0}$ determination

$x = {x}_{0} + \left({v}_{0} \cos \theta\right) t$
$y = {y}_{0} + \left({v}_{0} \sin \theta\right) t - \frac{1}{2} g {t}^{2}$

so the non-parametric orbit equation is

$y = {y}_{0} + \frac{{v}_{0} \sin \theta}{{v}_{0} \cos \theta} \left(x - {x}_{0}\right) - \frac{1}{2} g {\left(\frac{x - {x}_{0}}{{v}_{0} \cos \theta}\right)}^{2}$ or

$y = {y}_{0} + \tan \theta \left(x - {x}_{0}\right) - \frac{1}{2} g {\left(\frac{x - {x}_{0}}{{v}_{0} \cos \theta}\right)}^{2}$

now

$y = {y}_{0} + \tan \theta \left(x - {x}_{0}\right) - \frac{1}{2} g {\left(\frac{x - {x}_{0}}{{v}_{0} \cos \theta}\right)}^{2} = 0$

but ${x}_{0} = 0$ then we have

$y = {y}_{0} + \tan \theta x - \frac{g}{2} {\left(\frac{x}{{v}_{0} \cos \theta}\right)}^{2} = 0$ so

${y}_{0} + \tan \theta d - \frac{g}{2} {\left(\frac{d}{{v}_{0} \cos \theta}\right)}^{2} = 0$

with $d = 4$ [m]

Now solving for ${v}_{0}^{2}$

v_0^2 = g/2(d^2)/(cos^2theta(y_0+d tan theta)

with

$\theta = \frac{\pi}{6}$ [rad]
${y}_{0} = 1.2$ [m]
$g = 9.81$

giving ${v}_{0}^{2} = 29.82$

and finally

$K = 29.82$