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anonymous
 one year ago
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
anonymous
 one year ago
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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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0is this a physics question?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2from your equation of motion, I see that the initial speed is 36 feet/sec

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, the equation for speed at a generic time t, is: \[\Large v\left( t \right) = {v_0}  gt \to v\left( t \right) = 36  32t\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2in 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
 one year ago
Best ResponseYou've already chosen the best response.0No, Algebra @arindameducationusc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How am I supposed to do that.. can u help me please @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.2now, when the ball reaches its maximum height, its speed is zero, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2from your equation of h(t), I can say that the initial speed is 36 feet/sec, and not 32 feet/sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could 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
 one year ago
Best ResponseYou've already chosen the best response.0From a mathematical perspective, a) is asking for the tcoordinate of the vertex. This is given by \[t=\frac{ b }{ 2a }\]with b=36 and a=16 from the given function.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0none of the above https://gyazo.com/754cc35432c834269c8ad1f8315d4fbe

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0if this is algebra, you must be doing something like completing the square?? pls advise.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0quadratic equations and functions @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0so complete the square for \(h = –16t^2 + 36t + 9\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0helps 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
 one year ago
Best ResponseYou've already chosen the best response.0my 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{36}{32}=1.25\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that answers the first question a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to 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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