Answer:
[tex]t\approx 5.865\,s[/tex]
Step-by-step explanation:
The motion equations that describe the ball are, respectively:
[tex]x = \left[\left(152\,\frac{ft}{s} \right)\cdot \cos 52^{\circ} \right] \cdot t[/tex]
[tex]y = 4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}[/tex]
The time required for the ball to hit the ground is computed from the second equation. That is to say:
[tex]4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{m}{s^{2}} \right) \cdot t^{2} = 0[/tex]
Given that formula is a second-order polynomial, the roots of the equation are described below:
[tex]t_{1}\approx 5.865\,s[/tex] and [tex]t_{2} \approx -0.048\,s[/tex]
Just the first root offers a realistic solution. Then, [tex]t\approx 5.865\,s[/tex].
