How many feet?

[moronatmath]

Could I get some help with this problemo?

If a ball is thrown directly upward with a velocity of 44 feet per second, its height (in feet) after t seconds is given by y= 44 t -16 t^2. What is the maximum height attained by the ball?



Would I just plug in t=44 to find the maximum height?

If so is -29040 the correct anwser?

 

[tkhunny]

Forget that it is a ball. It's an equation of a parabola. These have very predictable behavior. The minimum or maximum is ALWAYS at the vertex. The vertex is ALWAYS equidistant from the roots.

Solve this 44*t - 16*t^2 = 0 to find the roots.

Show your work and quit guessing.

 

[moronatmath]

ok here is my work shwon for that
44t + -16t2 = 0
factor
4t(11 + -4t) = 0
t=0

than
11 + -4t = 0
11 + -11 + -4t = 0 + -11
0 + -4t = 0 + -11
-4t = -11
t = 2.75

 

[TchrWill]

Could I get some help with this problemo?

If a ball is thrown directly upward with a velocity of 44 feet per second, its height (in feet) after t seconds is given by y= 44 t -16 t^2. What is the maximum height attained by the ball?

Would I just plug in t=44 to find the maximum height?

If so is -29040 the correct anwser?

Alternatively:
The ball reaches its maximum height when the upward velocity is zero.
This point in time derives from the expression Vf = Vo - gt = Vo - 32t where Vo is the initial upward velocity, Vf is the final upward velocity, 0, and t = the time to Vf in seconds.
Therefore, Vf = 0 = 44 - 32t making t = 1.375 seconds.

Then, the maximum height becomes h = y = 44t - 16t^2 = 44(1.375) - 16(1.375)^2
or 60.5 - 30.25 = 30.25 feet.

 

[soroban]

Hello, moronatmath!

If a ball is thrown directly upward with a velocity of 44 feet per second,
its height (in feet) after t seconds is given by: \(\displaystyle \,y\:=\:44t\,-\,16 t^2\)
What is the maximum height attained by the ball?

Would I just plug in t=44 to find the maximum height?
Why use 44? \(\displaystyle \;\)Why not use the current Dow-Jones average?

If so, is -29040 the correct anwser?
\(\displaystyle \;\;\) Yeah, right . . . it's highest point is 5.5 miles underground!

Okay, enough sarcasm . . .

We have a parabola: \(\displaystyle \,y\:=\;-16t^2\,+\,44t\)
It is a down-opening parabola, so its vertex is the highest point.

Do you know how to find the vertex of a parabola?
\(\displaystyle \;\;\)The formula is: \(\displaystyle \,t\,=\;\frac{-b}{2a}\)

We have: \(\displaystyle \,a\,=\,-16,\;b\,=\,44\), so: \(\displaystyle \,t\,=\,\frac{-44}{2(-16)} \,=\,\frac{11}{8}\) seconds.

Then: \(\displaystyle \,y\:=\:-16\left(\frac{11}{8}\right)^2\,+\,44\left(\frac{11}{8}\right)\:=\:\frac{121}{4}\:=\:30.25\) feet.

 

[tkhunny]

ok here is my work shwon for that
44t + -16t2 = 0
factor
4t(11 + -4t) = 0
t=0

than
11 + -4t = 0
11 + -11 + -4t = 0 + -11
0 + -4t = 0 + -11
-4t = -11
t = 2.75

This is excellent work. Notice how Soroban's magic formula is right in front of you, just like I said. In between those two roots, (0+2.75)/2 = 1.375 = 11/8 -- This gives y = 30.25 ft.