Answer:
h = 112.5 feets
Step-by-step explanation:
The equation for horizontal distance "h" in feet of a projectile with initial velocity v₀ and initial angle theta is given by :
[tex]h=\dfrac{v_o^2}{16}\sin\theta\cos\theta[/tex]
We know that, [tex]2\sin\theta\cos\theta=\sin2\theta[/tex]
So,
[tex]h=\dfrac{v_o^2}{32}\sin2\theta[/tex]
Now we need to find the maximum horizontal distance possible and at what angle does this occur.
For maximum distance angle should be 45 degrees. Som,
[tex]h=\dfrac{60^2}{32}\sin2(45)\\\\h=112.5\ \text{feet}[/tex]
So, 112.5 feets is the maximum possible distance.
