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A ball travels on a parabolic path in which the height (in feet) is given by the expression -16t^2+80t+21, where t is the time after launch. What is the maximum height of the ball, in feet?

Dec 26, 2018

#1
+1

Factoring:: $$-16t^2+80t+21=t=-\frac{1}{4},\:t=\frac{21}{4}$$

We first factor $$-16t^2+80t+21$$ into $$-\left(4t+1\right)\left(4t-21\right)$$

Next, $$4t+1=0 \longrightarrow t=-\frac{1}{4}.$$

And, $$4t-21=0 \longrightarrow t=\frac{21}{4}.$$

Since the value has to be maximum, the answer is $$\boxed{\frac{21}{4}}.$$

Oh, I think I'm wrong....

.
Dec 26, 2018
edited by tertre  Dec 26, 2018
#2
+2

-16t^2 + 80t + 21

The  t  that maximizes this is given by :

-80 / [ 2 * -16]  =  80 / 32  =  5/2

Putting this back into the function, we get that the max height is

-16(5/2)^2 + 80(5/2) + 21  =

-16(25/4) + 200 + 21  =

-100 + 200 + 21 =

121 ft   Dec 26, 2018
#3
+1

Yes, you are correct CPhill. Misread the problem!

tertre  Dec 26, 2018
#4
+1

No big deal.....all that Christmas food has made my mind hazy, too....LOL!!!!   CPhill  Dec 26, 2018