Question

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

Answer 1

is this a physics question?

Answer 2

from your equation of motion, I see that the initial speed is 36 feet/sec

Answer 3

now, the equation for speed at a generic time t, is:
\[\Large v\left( t \right) = {v_0} - gt \to v\left( t \right) = 36 - 32t\]

Answer 4

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

Answer 5

No, Algebra @arindameducationusc

Answer 6

How am I supposed to do that.. can u help me please @Michele_Laino

Answer 7

the general formulas of kinematics, are:
\[\Large \begin{gathered}
h\left( t \right) = - \frac{1}{2}g{t^2} + {v_0}t + {h_0} \hfill \\
v\left( t \right) = {v_0} - gt \hfill \\
\end{gathered} \]

Answer 8

now, when the ball reaches its maximum height, its speed is zero, right?

Answer 9

from your equation of h(t), I can say that the initial speed is 36 feet/sec, and not 32 feet/sec

Answer 10

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?

Answer 11

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.

Answer 12

none of the above

https://gyazo.com/754cc35432c834269c8ad1f8315d4fbe

Answer 13

if this is algebra, you must be doing something like completing the square??

pls advise.

Answer 14

quadratic equations and functions @IrishBoy123

Answer 15

so

complete the square for \(h = –16t^2 + 36t + 9\)

Answer 16

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...

Answer 17

oh my

Answer 18

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 \[-\frac{36}{2\times (-16)}=1\]

Answer 19

ok not 1 sorry

Answer 20

\[\frac{36}{32}=1.25\]

Answer 21

that answers the first question
a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

Answer 22

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
\[h=-16t^2+32t+9\]