Projectile launcher problem?

Question

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.

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Answer 1 · 2018-03-01T16:53:07

$29.82$

Let $E_k = 1/2 m v_0^2$ be the ball kinetic energy leaving the barrel. Then

$1/2K (Delta x)^2 = 1/2 m v_0^2$ so

$K = m(v_0/(Delta x))^2$

where

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

Follows the $v_0$ determination

$x = x_0 + (v_0 cos theta)t$
$y = y_0 + (v_0 sin theta) t - 1/2 g t^2$

so the non-parametric orbit equation is

$y = y_0 +(v_0 sin theta)/(v_0 cos theta)(x-x_0)-1/2 g ((x-x_0)/(v_0 cos theta))^2$ or

$y = y_0 +tan theta(x-x_0)-1/2 g ((x-x_0)/(v_0 cos theta))^2$

now

$y =y_0 +tan theta(x-x_0)-1/2 g ((x-x_0)/(v_0 cos theta))^2=0$

but $x_0 = 0$ then we have

$y = y_0 + tan theta x -g/2 (x/(v_0 cos theta))^2=0$ so

$y_0 + tan theta d -g/2 (d/(v_0 cos theta))^2=0$

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

$theta = pi/6$ [rad]
$y_0 = 1.2$ [m]
$g = 9.81$

giving $v_0^2 = 29.82$

and finally

$K = 29.82$