Answer:
[tex]v_o=40.14\ m/s[/tex]
Explanation:
Horizontal Launch
It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:
We have the following formulas
[tex]\displaystyle v_x=v_o[/tex]
[tex]\displaystyle x=v_o.t[/tex]
[tex]\displaystyle v_y=g.t[/tex]
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
Where [tex]v_o[/tex] is the initial horizontal speed, [tex]v_y[/tex] is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.
If we know the initial height of the object, we can compute the time it takes to hit the ground by using
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
Rearranging and solving for t
[tex]\displaystyle 2y=gt^2[/tex]
[tex]\displaystyle t^2=\frac{2\ y}{g}[/tex]
[tex]\displaystyle t=\sqrt{\frac{2\ y}{g}}[/tex]
We then replace this value in
[tex]\displaystyle x=v_o.t[/tex]
To get
[tex]\displaystyle v_o=\frac{x}{t}[/tex]
[tex]\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}[/tex]
[tex]\displaystyle v_o=\sqrt{\frac{g}{2y}}.x[/tex]
The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing [tex]v_o[/tex]
[tex]\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6[/tex]
The launch velocity is
[tex]\boxed{v_o=40.14\ m/s}[/tex]
