anonymous
  • anonymous
A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 36t + 9. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball's maximum height? A) 1 s; 22 ft B) 2 s; 22 ft C) 2 s; 6 ft D) 1 s; 54 ft
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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arindameducationusc
  • arindameducationusc
is this a physics question?
Michele_Laino
  • Michele_Laino
from your equation of motion, I see that the initial speed is 36 feet/sec
Michele_Laino
  • Michele_Laino
now, the equation for speed at a generic time t, is: \[\Large v\left( t \right) = {v_0} - gt \to v\left( t \right) = 36 - 32t\]

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More answers

Michele_Laino
  • Michele_Laino
in order to compute the time t, at which the ball has reached its maximum height, you have to set v=0, so you have to solve this equation: \[\Large 0 = 36 - 32t\] please solve for t
anonymous
  • anonymous
No, Algebra @arindameducationusc
anonymous
  • anonymous
How am I supposed to do that.. can u help me please @Michele_Laino
Michele_Laino
  • Michele_Laino
the general formulas of kinematics, are: \[\Large \begin{gathered} h\left( t \right) = - \frac{1}{2}g{t^2} + {v_0}t + {h_0} \hfill \\ v\left( t \right) = {v_0} - gt \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now, when the ball reaches its maximum height, its speed is zero, right?
Michele_Laino
  • Michele_Laino
from your equation of h(t), I can say that the initial speed is 36 feet/sec, and not 32 feet/sec
anonymous
  • anonymous
Could there not also be a horizontal component of initial velocity such that the vertical component is 32 m/s yet and horizontal component is such that initial velocity is 36 m/s?
anonymous
  • anonymous
From a mathematical perspective, a) is asking for the t-coordinate of the vertex. This is given by \[t=-\frac{ b }{ 2a }\]with b=36 and a=-16 from the given function.
IrishBoy123
  • IrishBoy123
none of the above https://gyazo.com/754cc35432c834269c8ad1f8315d4fbe
IrishBoy123
  • IrishBoy123
if this is algebra, you must be doing something like completing the square?? pls advise.
anonymous
  • anonymous
quadratic equations and functions @IrishBoy123
IrishBoy123
  • IrishBoy123
so complete the square for \(h = –16t^2 + 36t + 9\)
IrishBoy123
  • IrishBoy123
helps to take the leading coefficient outside, as in \(\large h(t)=–16(t^2-\frac{36}{16}t-\frac{9}{16})\) lots of square numbers in there so should be fine...
anonymous
  • anonymous
oh my
anonymous
  • anonymous
my guess is that this has nothing to do with calculus it is really not this complicated \[ h = –16t^2 + 36t + 9 \] is a parabola that opens down the maximum is the second coordinate of the vertex the first coordinate of the vertex of \[y=ax^2+bx+c\] is always \[-\frac{b}{2a}\] which, in our case is \[-\frac{36}{2\times (-16)}=1\]
anonymous
  • anonymous
ok not 1 sorry
anonymous
  • anonymous
\[\frac{36}{32}=1.25\]
anonymous
  • anonymous
that answers the first question a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.
anonymous
  • anonymous
to find the maximum height replace \(t\) by \(1.25\) i see your answer choices do not have \(1.25\) so either there is a mistake in the answers, or it should have been \[h=-16t^2+32t+9\]

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