Question

A tennis player hits a ball 3 feet above the ground with a velocity of 60 ft/sec. After how many seconds will the ball be at a height of 3 feet again?

Answer 1

Answer:

The ball will be at a height of 3 feet above the ground after 3.729 seconds.

Step-by-step explanation:

Let suppose that the tennis player has hit the ball vertically, meaning that ball will experiment a free fall, that is, an uniform accelerated motion due to gravity. The time taken by the ball in terms of its initial velocity ($v_{o}$), in feet per second, initial position ($x_{o}$), in feet, final position ($x$), in feet, and gravitational acceleration ($g$), in feet per square second is described by this second order polynomial:

$x = x_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g\cdot t^{2}$ (1)

If we know that $x_{o} = x = 3\,ft$,  $v_{o} = 60\,\frac{ft}{s}$ and $g = -32.174\,\frac{ft}{s^{2}}$, then the time needed by the ball to be at a height of 3 feet again is:

$3\,ft = 3\,ft + \left(60\,\frac{ft}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right)\cdot t^{2}$

$-16.087\cdot t^{2}+60\cdot t = 0$ (2)

$t^{2} - 3.729\cdot t = 0$

$t \cdot (t -3.729) = 0$

The binomial contains the time when the ball will be at a height of 3 feet again. In other words, the ball will be at a height of 3 feet above the ground after 3.729 seconds.