SOLUTION: An object has been thrown straight up into the air. The formula
h = vt − 16t2
gives the height h of the object above the ground after t seconds, when it is thrown upward with
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Question 1162345: An object has been thrown straight up into the air. The formula
h = vt − 16t2
gives the height h of the object above the ground after t seconds, when it is thrown upward with an initial velocity v.
Before a football game, a coin toss is used to determine which team will kick off. The height h (in feet) of a coin above the ground t seconds after being flipped up into the air is given by h = −16t2 + 70t + 9
How long will the coin be in the air?
Found 3 solutions by ikleyn, Alan3354, solver91311:
Answer by ikleyn(52832) (Show Source): You can put this solution on YOUR website!
.
Notice that your "general" formula at the beginning of the post
does not correlate with the formula at the end.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
h = vt − 16t2
---------
Use ^, Shift 6, for exponents.
eg, 16t^2 for
=============================
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The general function for the height of a projectile launched vertically as a function of time is:
\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ h_o)
Where 
is the acceleration resulting from the force of gravity, either 
or 
depending on whether you are using fps or mks units, 
is the initial velocity, 
is the height at launch, and 
is the elapsed time in seconds since launch.
Given your lead coefficient of -16, I'll assume fps.
Assuming the referee allowed the count to hit the ground rather than catching it as is typical behavior at the start of football games, then the coin will be in the air from the time of launch until it hits the ground. When it hits the ground its height above the ground is necessarily zero.
So solve:
Note that since the initial height is more than zero, the two roots of this equation must be of opposite signs. You want the positive answer because what happened before he threw the coin is moot.
As a side note, if the referee was able to reach 9 feet in the air standing flat-footed, he would have to be about 6 feet 9 inches tall on average. 6 feet would be a more reasonable value for initial height. Who reaches above their head to release the coin when tossing it?
John
My calculator said it, I believe it, that settles it
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