The function #h(t)=-16t^2+80t# represents the height of the baseball over time. How long do you think the ball will be in the air?

Answer 1

#5# time units.

This is not a question for calculus!

#h(t) = 0# at ground level
#h(t) = 0 => -16t^2+80t=0 # #:. 16t^2 - 80t=0 # #:. 16t(1-5)=0 # #t=0 #, or #t=5#
Interpretation #t=0# represents the time at which the baseball first left the ground #t=5# represents the time at which the baseball returns to the ground
So the ball is in the air for #5# time units.
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Answer 2

To find out how long the ball will be in the air, we need to find the time when the height function ( h(t) ) equals zero, because at that point, the ball hits the ground. This can be done by solving the equation ( h(t) = 0 ) for ( t ). By factoring, we get:

[ h(t) = -16t^2 + 80t ] [ -16t(t - 5) = 0 ]

This equation is true when either ( t = 0 ) or ( t - 5 = 0 ). So, the two possible solutions are ( t = 0 ) and ( t = 5 ) seconds. Since ( t = 0 ) represents the starting time when the ball is thrown, we consider the positive solution ( t = 5 ) seconds as the time when the ball will be in the air.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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