Question

1.

Answer 1

Answer: 2.61 s

Explanation:

We are given the following data:

$s=5 ft$ is the initial height of the object

$v=40 ft/s$ is the initial height of the object

In addition, the motion of the object is given by:

$h=-16t^{2}+ vt +s$ (1)

Where:

$h=0 ft$ is the final height of the object

$t$ is the time the object is in the air before hitting the ground

Rewritting (1) with the given data:

$0=-16t^{2}+ 40t +5$ (2)

Solving the quadratic equation with the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=-16$, $b=40$, $c=5$ and choosing the positive value of time:

$t=\frac{-40\pm\sqrt{(-40)^{2}-4(-16)(5)}}{2(-16)}$ (3)

$t=2.61 s$ (4) This is the time