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A ball travels on a parabolic path in which the height (in feet) is given by the expression $-25t^2+75t+24$, where $t$ is the time after launch. At what time is the height of the ball at its maximum?

 Nov 4, 2014

Best Answer 

 #4
avatar+118703 
+5

gino has just replaced all the t's with x's.

He has only done this because it is much mor normal to see equations like this with an x  instead of a t.

There are no x'es in your question.  

height=25t2+75t+24$Theaxisofsymmetrywillbeattime$t=75225=t=32=1.5secs

 

So the ball will be at the maximum height after 1.5 seconds

 Nov 4, 2014
 #1
avatar+23254 
+5

You are looking for the x-value of the vertex.

An easy way to find this point is to graph it.

If you do not want to graph it, then memorize the formula for finding the x-value of the vertex:  x  =  -b/(2a).

Because your expression is written correctly,  a = -25,  b = 75,  and  c = 24.

Therefore, the x-value of the vertex is  -(75) / (2·-25)  =  -75 / -50  =  1.5   <--- the time.

To find the height, replace x wby 1.5 in the original expression.

 Nov 4, 2014
 #2
avatar
0

Where do you find x? Can you hel me find the answer?

 Nov 4, 2014
 #3
avatar+33657 
0

Deleted

 

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 Nov 4, 2014
 #4
avatar+118703 
+5
Best Answer

gino has just replaced all the t's with x's.

He has only done this because it is much mor normal to see equations like this with an x  instead of a t.

There are no x'es in your question.  

height=25t2+75t+24$Theaxisofsymmetrywillbeattime$t=75225=t=32=1.5secs

 

So the ball will be at the maximum height after 1.5 seconds

Melody Nov 4, 2014

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