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.)

Answer 1

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of free fall objects

Our knowable variables are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the rock hits the ground making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we evaluate the equation:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of Quadratic Formula, we can solve it using:

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

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the time can not be negative, the reasonable answer is

$t=6.4534s$