Use the model for projectile motion, assuming there is no air resistance and g = 32 feet per second per second.
A bale ejector consists of two variable-speed belts at the end of a baler. Its purpose is to toss bales into a trailing wagon. In loading the back of a wagon, a bale must be thrown to a position 8 feet above and 16 feet behind the ejector. Find the minimum initial speed of the bale, v, and the corresponding angle, θ, at which it must be ejected from the baler. (Round your answers to two decimal places.)

Let [tex]v_v, v_h[/tex] be the vertical and horizontal components required of the initial velocity of the bale. For the bale to be shot to 8ft above with acceleration of g = -32 ft/s2, we can calculate it using the following equation of motion:

[tex]v^2 - v_v^2 = 2a\Delta s[/tex]

where v = 0 m/s is the final velocity of the bale when reaches the trailing wagon, [tex]v_v[/tex] is the initial vertical velocity of the bale, g = -32 m/s2 is the deceleration of the bale, and [tex]\Delta h[/tex] is the vertical distance traveled.

[tex]0 - v_v^2 = 2*(-32)*8 = -512 [/tex]

[tex]v_v = \sqrt{512} = 22.6 m/s[/tex]

The time it takes to travel that distance can be calculated using the following equation of motion

[tex]s = v_vt + gt^2/2[/tex]

[tex]8 = 22.6t - 32t^2/2[/tex]

[tex]16t^2 - 22.6t + 8 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{22.6\pm \sqrt{(-22.6)^2 - 4*(16)*(8)}}{2*(16)}[/tex]

[tex]t= \frac{22.6\pm0}{32}[/tex]

t = 0.707s

This is also the time it takes for the bale to travel 16ft horizontally. So we can calculate the initial horizontal velocity

[tex]v_h = 16 / 0.707 = 22.63 m/s[/tex]

So the magnitude and direction of the whole initial velocity is

[tex]v = \sqrt{v_v^2 + v_h^2} = \sqrt{22.6^2 + 22.63^2} = \sqrt{510.76 + 512.1169} = \sqrt{1022.8769} = 31.98 m/s[/tex]

[tex]tan\theta = \frac{v_v}{v_h} = \frac{22.6}{22.63} = 1[/tex]

[tex]\alpha = tan^{-1}1 = 0.78 rad \approx 44.96^0 [/tex]