Question

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at that time, y0 is the initial height (at t=0), v0 is the initial velocity, and g=9.8m/s2 (the acceleration due to gravity). If you stand at the edge of a cliff that is 75 m high and throw a rock directly up into the air with a velocity of 20 m/s, at what time will the rock hit the ground? (Note: The Quadratic Formula will give two answers, but only one of them is reasonable.) View Available Hint(s)

Answer 1

Answer: 6.45 s

Explanation:

We have the following equation:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height when the rock hits the ground

$y_{o}=75 m$ the height at the edge of the cilff

$V_{o}=20 m/s$ the initial velocity

$g=9.8 m/s^{2}$ acceleration due gravity

$t$ time

$0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}$  (2)

Rearranging the equation:

$-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0$ (3)

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$, and we have to use the quadratic formula if we want to find  $t$:

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$  (4)

Where $a=-4.9$, $b=20$, $c=75$

Substituting the known values and choosing the positive result of the equation:

$t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}$  (5)

$t=6.453 s$  This is the time it takes to the rock to hit the ground