The height above the ground of a ball thrown up with velocity of 96 feet per second from a height of 6 feet is 6+96t-16t^2 feet, where t is the time in seconds. According to this model, how high is the ball after 7 seconds? Explain.
I substituted in for t, but I got a negative answer.

Question

anonymous · Answer

heightt =  6+ 96t - 16t^2

so if t = 7
heightt =  6+ 96t - 16t^2
heightt =  6+ 96(x7) - 16(x7^2)
heightt =  6+ 96(x7) - 16(x49)
heightt =  ??? @Askswatw260 ?

askswatw260 · Answer

what do you mean???? @Jack4

whpalmer4 · Answer

No, you plug in t = 7, not t = x7
$$h(t) = 6+96t - 16t^2$$$$h(7) = 6+96(7)-16(7)^2 = $$

anonymous · Answer

-106, right?

askswatw260 · Answer

yeah I got that but how is that so? @Jack4

whpalmer4 · Answer

yes, -106 is the number returned.

A plot of the function from t = 0 to t = 7 may help understand what is going on here...

askswatw260 · Answer

for the worksheet how would I explain -106?

whpalmer4 · Answer

Well, first, do you understand what the 3 terms in that equation represent?

askswatw260 · Answer

I pretty sure I do

anonymous · Answer

it's because you consider the speed it's thrown at to be positive...
have a look at drawing to follow

askswatw260 · Answer

thank you @Jack4 and @whpalmer4

anonymous · Answer

|dw:1396733382338:dw|

anonymous · Answer

all good dude

whpalmer4 · Answer

The reason you get a negative number is that the equation doesn't have any knowledge of where the ground actually is, and that the ball will stop there.  t=7 seconds is a time beyond the time of impact.  The equation is a model which only applies to the domain $t = 0 \text{ to }t\approx 6.062$