Physics: 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.)

An object moving due to gravity can be described

DestinyK8033

02.01.2022

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

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 :

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

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