Question

A baseball player hits a ball at an angle of 52 degrees and at a height of 4.5 ft. if the ball's initial velocity after being hit is 152ft/s and if no one catches the ball when will it hit the ground?

Answer 1

Answer:

$t\approx 5.865\,s$

Step-by-step explanation:

The motion equations that describe the ball are, respectively:

$x = \left[\left(152\,\frac{ft}{s} \right)\cdot \cos 52^{\circ} \right] \cdot t$

$y = 4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

The time required for the ball to hit the ground is computed from the second equation. That is to say:

$4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{m}{s^{2}} \right) \cdot t^{2} = 0$

Given that formula is a second-order polynomial, the roots of the equation are described below:

$t_{1}\approx 5.865\,s$ and $t_{2} \approx -0.048\,s$

Just the first root offers a realistic solution. Then, $t\approx 5.865\,s$.