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

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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
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is this a physics question?
from your equation of motion, I see that the initial speed is 36 feet/sec
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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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
No, Algebra @arindameducationusc
How am I supposed to do that.. can u help me please @Michele_Laino
the general formulas of kinematics, are: \[\Large \begin{gathered} h\left( t \right) = - \frac{1}{2}g{t^2} + {v_0}t + {h_0} \hfill \\ v\left( t \right) = {v_0} - gt \hfill \\ \end{gathered} \]
now, when the ball reaches its maximum height, its speed is zero, right?
from your equation of h(t), I can say that the initial speed is 36 feet/sec, and not 32 feet/sec
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?
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.
none of the above https://gyazo.com/754cc35432c834269c8ad1f8315d4fbe
if this is algebra, you must be doing something like completing the square?? pls advise.
quadratic equations and functions @IrishBoy123
so complete the square for \(h = –16t^2 + 36t + 9\)
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...
oh my
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 \[-\frac{36}{2\times (-16)}=1\]
ok not 1 sorry
\[\frac{36}{32}=1.25\]
that answers the first question a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.
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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