<span>adjacent means close too</span>
So much clearer! Thank you (seriously).
Use the point-slope form of the eq'n of a str. line. Find the slope first:
0-(-2.5) 2.5
m = ------------- = -------- = 1/2
5 - 0 5
Then, y-0 = (1/2)(x-5)
Which answer choice do you believe matches this?
I see that you are in high school, and I'm hoping that you've been introduced
to differential calculus, because I don't know how to answer this question without
using it.
We're told that Jason's height above the water is <em>H(t) = -16t² + 16t + 480 .</em>
We can observe many things from this equation:
-- Up is the positive direction; down is the negative direction.
-- The acceleration of gravity is 32 ft/sec² .
-- Jason jumps upward from the cliff, at 16 ft/sec .
-- The cliff is 480-ft above the water.
(This tells us why the question is only concerned with his maximum height,
and then it ends ... 480-ft is one serious cliff, and what happens after the
peak of his arc is too gruesome to contemplate.)
In any case, his vertical velocity is the first derivative, with respect to time,
of his height above the water.
V = -32 t + 16
At the peak of his arc, where gravity takes over, his velocity changes from
upward to downward, and it's momentarily zero.
0 = -32t + 16
Add 32t to each side: 32t = 16
Divide each side by 32: <em> t = 1/2 second</em>
His height at that instant is H(0.5) = -16(0.5)² + 16(0.5) + 480 =
<em>4-ft above the cliff, 484-ft above the water</em>,
and then he begins falling from that altitude.
The duration of his dive is 484 = 16 t²
t = √(484/16) = <em>5.5 seconds</em>
and he hits the water at V = a t = (32) x (5.5) = 176 ft/sec = <em>exactly 120 mph </em>
Jason was good man ... a good student, and always kind to everyone he met.
He will certainly be missed.
The cross-section is a rectangle. Its height is the same as the height of the prism, while its length is not necessarily the same but certainly perpendicular to the height.
I think this might be the Answer
