SOLUTION: The height of a ball launched from the top of a hill, where t is time, is (5t-3)(6-t)=h. If h=0, when the ball hits the ground, how long does it take for the ball to hit the ground
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Question 439691: The height of a ball launched from the top of a hill, where t is time, is (5t-3)(6-t)=h. If h=0, when the ball hits the ground, how long does it take for the ball to hit the ground?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Your function makes no sense at all in the physical world.
Solving it for t when h = 0 yields 3/5 seconds and 6 seconds. How can something launched from the top of a hill reach 0 height twice?
Furthermore, if you multiply the two binomials you get
Though you failed to mention what units the height is measured in.
The function for the height of a projectile launched near the surface of any celestial body is given by:
\ =\ -\frac{1}{2}g_it^2\ +\ v_ot\ +\ h_o)
Where 
is the acceleration due to gravity of the selected celestial body, 
or 
, in the case of Earth, 
is the initial velocity, and 
is the initial height.
If you assume that you are using the mks system, then your gravitational acceleration, assuming you took very inappropriate roundoff liberties would allow for being on Earth. I don't know where you are if you are using fps.
But the problem is the anomaly caused by the fact that you are claiming to have launched from the top of a hill, but your initial height is a negative number. Granted, I am only 62 years old, so my experience is somewhat limited, but every hill in my experience has had a positive value for the height.
John
My calculator said it, I believe it, that settles it
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