Question

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.

Answer 1

`29.82`

Explanation:

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`