Answer:
1.065 km
Explanation:
The horizontal and vertical component of the canon ball when it's fired is
[tex]v_h = vcos\alpha = 120cos23^0 = 110.46 m/s[/tex]
[tex]v_v = vsin\alpha = 120sin23^0 = 46.89 m/s[/tex]
If we ignore air resistance, then gravitational acceleration g = -9.8m/s2 is the only thing that affects the vertical motion of the cannon ball. We can use the following equation to calculate the time it stays on air after it traveling h = -3.5m (down below the firing point) vertically
[tex]h = v_vt + gt^2/2[/tex]
[tex]-3.5 = 46.89t - 9.8t^2/2[/tex]
[tex]-4.9t^2 + 46.89t + 3.5 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{-46.89\pm \sqrt{(46.89)^2 - 4*(-4.9)*(3.5)}}{2*(-4.9)}[/tex]
[tex]t= \frac{-46.89\pm47.62}{-9.8}[/tex]
t = -0.07 or t = 9.64
Since t can only be positive we will pick t = 9.64
This is also the time it takes for the cannon ball to travel horizontally at the rate of 110.46 m/s
[tex]s = v_ht = 110.46 * 9.64 = 1065 m[/tex] or 1.065 km
