Check the picture below.
so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?
now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.
![\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20feet%7D%20%5C%5C%5C%5C%20h%28x%29%20%3D%20-16x%5E2%2Bv_ox%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%26%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%5C%5C%20x%3D%5Ctextit%7Bseconds%7D%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%200%3D-16%281%29%5E2%2B0x%2Bh_o%5Cimplies%200%3D-16%2Bh_o%5Cimplies%2016%3Dh_o%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%20%3D%20-16x%5E2%2B16~%5Chfill)
quick info:
in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².
The formula is for slope is y2-y1/x2-x1 so 3- -1 = 4 and -2 - 4 = -6 so it would be 4/-6
Answer:
Step-by-step explanation:
Remark
You could do this by converting the distance but the much more common way is to use kg and m.
Formula
PE = m * g * h
Givens
m = 40000 grams
m = 40000 grams * 1kg/1000 grams = 40 kg
g = 9.81
h = 35 m
Solution
PE = 40 * 9.81 * 35
PE = 13734 Joules
We know that the Pythagorean Theorem is a² + b² = c² and that the area of a square is l x w.
Firstly, we'll have to find the two measures of the triangle that correspond to the areas.
Since the figures are squares, we know that the length and width values must be the same.
We could square the numbers to find the side lengths, however we would have to square them again when substituting for the Pythagorean Theorem, so we can leave them as-is and adjust the equation accordingly.
(33) + b² = (44)
Next, we'll subtract our smaller value from our larger.
b² = (11)
Once again, we could find the square root of this number, but we'd just have to square it again to find the area of the square, so we can just simply write our answer as 11 units.
Therefore, the area of the square is 11 units!
<em>Hope this helped! :)</em>
