SOLUTION: A ball is thrown into the air. Its height (in feet) t seconds later is given by
h(t)=96t−16t2
Based on the two equivalent forms of the function h(t)=96t−16t2=t⋅
Algebra.Com
Question 1121393: A ball is thrown into the air. Its height (in feet) t seconds later is given by
h(t)=96t−16t2
Based on the two equivalent forms of the function h(t)=96t−16t2=t⋅(96−16t), answer the following questions:
a) What is the height of the ball 1 second after it has been thrown?
(include help (units) in your answer)
b) After how many seconds does the ball hit the ground?
(include help (units) in your answer)
c) At what time(s) is the ball 40 feet above the ground? If there is more than one answer, give your answer as a comma separated list of values. Note that you do not have to include units in this answer.
After
seconds
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Presuming that the person who threw the ball was standing in a hole of such depth that when the ball was released it was exactly at ground level, then your function makes sense.
\ =\ 96t\ -\ 16t^2)
1. Evaluate )
.
2. Set \ =\ 0)
, that is \ =\ 0)
and then solve for 
. Discard the zero root because 0 is the time the ball was thrown.
3. Set \ =\ 40)
, that is \ =\ 40)
and then solve for . There will be two roots, one representing the time the ball was at 40 feet from the ground on the way up, and one representing the time it was at 40 feet from the ground on the way down.
Extra credit: What is the highest point that the ball reaches? In other words, does it make sense that the ball was at 40 ft above the ground twice during its travels? Did it ever get to 40 ft high at all?
Hint: Put your function into \ =\ at^2\ +\ bt\ +\ c)
form, then calculate 
and evaluate the function at that value.
John
My calculator said it, I believe it, that settles it
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