<span>H(t) = −16t^2 + vt + s
</span>
<span>H(t) = −16t^2 + 100t + 140
H(t) = 0 = </span><span>−16t^2 + 100t + 140
16t^2 - 100t -140 = 0
t = [100 +/- √ (100^2 -4(16)(-140) ) ] /(2(16))
t = 7.43 and t = - 1.18.
El que tiene sentido para la pregunta es t = 7.43 s.
Option B) 7 s (approximately)
</span>
Answer:
61 and -87
Step-by-step explanation:
If the numbers are x and x - 148, we can write the following equation:
x + x - 148 = -26
2x - 148 = -26
2x = 122
x = 61 so x - 148 = 61 - 148 = -87
Answer what? This question? Oh, I'm answering it right now. Did you have another question out? Well, I can't see it.
17.4736843 or 17 remainder 45
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
