We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
196
4
avatar+80 

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
avatar+4322 
+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
avatar+103131 
+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

 

 

cool cool laugh

 Dec 26, 2018
 #3
avatar+4322 
+1

Yes, you are correct CPhill. Misread the problem!

tertre  Dec 26, 2018
 #4
avatar+103131 
+1

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

 

 

cool cool cool

CPhill  Dec 26, 2018

6 Online Users

avatar