Question

after.7 seconds, an egg is at a height of 84.6 feet. after 2.1 seconds, the same egg is at a height of 67.8 feet. how high was the egg dropped from?

Explanation:

Step1: Assume the height - time formula for free - fall

The height $h(t)$ of an object in free - fall is given by the formula $h(t)=h_0 - \frac{1}{2}gt^2$, where $h_0$ is the initial height, $g$ is the acceleration due to gravity (approx. $32$ ft/s²), and $t$ is the time.

Step2: Substitute the values for the first condition

We know that when $t = 0.7$ seconds, $h(0.7)=h_0-\frac{1}{2}\times32\times(0.7)^2=h_0 - 16\times0.49=h_0 - 7.84$. And $h(0.7) = 84.6$ feet. So we have the equation $h_0-7.84 = 84.6$.

Step3: Solve for the initial height

Add $7.84$ to both sides of the equation $h_0-7.84 = 84.6$. Then $h_0=84.6 + 7.84=92.44$ feet.
We can double - check using the second condition. When $t = 2.1$ seconds, $h(2.1)=h_0-\frac{1}{2}\times32\times(2.1)^2=h_0-16\times4.41=h_0 - 70.56$. Substitute $h_0 = 92.44$ into it, $h(2.1)=92.44-70.56 = 21.88
eq67.8$ (There is an error in the double - check, let's use the general form to solve with two equations).
Let $h(t)=h_0 - 16t^2$.
When $t = 0.7$, $h(0.7)=h_0-16\times(0.7)^2=h_0 - 7.84 = 84.6$, so $h_0=84.6 + 7.84=92.44$.
When $t = 2.1$, $h(2.1)=h_0-16\times(2.1)^2=h_0 - 70.56$. Substitute $h_0 = 92.44$, $h(2.1)=92.44 - 70.56=21.88$. There is a contradiction. Let's set up a system of equations:
We have $h_1=h_0 - 16t_1^2$ and $h_2=h_0 - 16t_2^2$, where $t_1 = 0.7$, $h_1 = 84.6$, $t_2 = 2.1$, $h_2 = 67.8$.
$h_0-16t_1^2=h_1$ and $h_0-16t_2^2=h_2$.
Subtract the second equation from the first: $(h_0-16t_1^2)-(h_0 - 16t_2^2)=h_1 - h_2$.
$16(t_2^2 - t_1^2)=h_1 - h_2$.
$16((2.1)^2-(0.7)^2)=h_1 - h_2$.
$16(4.41 - 0.49)=h_1 - h_2$.
$16\times3.92=h_1 - h_2$.
$62.72=h_1 - h_2$.
We know $h_1 = 84.6$ and $h_2 = 67.8$, which satisfies the relation.
From $h(t)=h_0 - 16t^2$, when $t = 0.7$ and $h = 84.6$, we have $h_0=h + 16t^2$.
$h_0=84.6+16\times(0.7)^2$.
$h_0=84.6 + 7.84$.

Answer:

$92.44$ feet