Question

Maria throws a softball straight up into the air with an initial velocity of 50 ft/sec. The ball leaves her hand when it is exactly 4 ft. Above the ground. Write an equation that models the situation.

Answer 1

Answer:

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$.

Step-by-step explanation:

Let suppose that effects from air friction and Earth's rotation can be neglected, so that the softball can be modelled experiment a free fall, that is, an uniform accelerated motion due to gravity. From we know the initial velocity and position of the position and we can determine the final position of the ball as a function of the final velocity:

$v_{f}^{2} = v_{o}^{2}+2\cdot a \cdot (s_{f}-s_{o})$ (1)

Where:

$s_{o}$, $s_{f}$ - Initial and final position of the softball, measured in feet.

$a$ - Acceleration, measured in meters per square second.

$v_{o}$, $v_{f}$ - Initial and final velocities of the softball, measured in feet per second.

If we know that $v_{o} = 50\,\frac{ft}{s}$, $a = -32.174\,\frac{m}{s^{2}}$ and $s_{o} = 4\,ft$, then the equation that models the situation is:

$v_{f}^{2} = 2500-64.348\cdot (s_{f}-4)$

Then, we clear the final position of the softball:

$-\frac{v_{f}^{2}-2500}{-64.348} = s_{f}-4$

$s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$.