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Question

A ball is thrown from a height of
217
feet with an initial downward velocity of
17/fts
. The ball's height
h
(in feet) after
t
seconds is given by the following.
=h−217−17t16t2
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

Assuming that the equation is h = 217 - 17t - 16t²:

 

We will let h = 0:

 

Why? It actually depends on the equation. If you observe when t = 0, (substituting it to the equation) you will get h = 217 which is the initial height, in time where nothing is happening yet. Since you threw it down, that initial height will go down as well up to the ground which is h = 0.

Being h = 0; you will have:

 

0 = 217 - 17t - 16t² (which is a quadratic equation)

 

Using quadratic formula, you will get:

 

t = 3.19 and -4.25.

 

True, there are two answers. However, we are talking about time here and there's no such thing as negative time. Making the -4.25 an ERRONEOUS SOLUTION. Leaving the 3.19 seconds as the only correct solution.

 

t = 3.19 seconds