An arrow is launched upward with a velocity of 192 feet per second from the top of a 95-foot stage. What is the maximum height a
ttained by the arrow?
1 answer:
9514 1404 393
Answer:
671 feet
Step-by-step explanation:
There are a couple of ways to figure this. One is to use a sort of shortcut equation to find the distance traveled (d) by an object when subject to some initial velocity (v) and acceleration (a). Here the acceleration due to gravity is -32 ft/s².
v² = 2ad
d = v²/(2a) = (192 ft/s)^2/(2·32 ft/s²) = 576 ft
This height is in addition to the starting height of 95 ft, so the arrow's maximum height is ...
max height = 95 ft + 576 ft = 671 ft
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Another way to work this problem is to start with the equation for ballistic motion. Filling in the given initial velocity and height, we have ...
h(t) = -16t^2 +192t +95
The time the arrow reaches the maximum height is the time representing the axis of symmetry of the parabola:
t = -(192)/(2(-16)) = 6
Then the maximum height is ...
h(6) = -16·6^2 +192·6 +95 = 671
The maximum height is 671 feet.
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<em>Additional comment</em>
For the standard-form quadratic ...
y = ax^2 +bx +c
The axis of symmetry is ...
x = -b/(2a)
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Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as: