Question 339834: an object is dropped from an initial height of x feet. the objects height at any time t. on seconds , is given by h=-16t^2+s. How long does it take for an object dropped from 300 feet to hit the ground ?
Found 2 solutions by nerdybill, solver91311:
Answer by nerdybill(7384)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
If you are going to drop your object from an initial height of  feet, then you need to use the function:
\ =\ -16t^2\ +\ x)
On the other hand, if you insist on using
\ =\ -16t^2\ +\ s)
then you must drop your object from an initial height of  feet.
In point of fact, it will serve you well to learn the more general function for the height of an object in feet acted upon by gravitational force near the surface of planet Earth.
\ =\ -16t^2\ +\ v_ot\ +\ h_o)
Where  is time in seconds,  is initial velocity in feet per second, and  is the initial height in feet.  is the acceleration due to gravity near the Earth in feet per second per second divided by 2.
For your problem:  (the object was merely dropped, not thrown),  , \ =\ 0) (since we want to know the time when the object hits the ground, i.e. height = zero), and is what we want to determine.
Plug in the values we know:
t\ +\ 300)
A little algebra:
Now all you need to do is solve the quadratic for the positive root (We can't go back in time, now can we?). Let me know what you find out.
John
My calculator said it, I believe it, that settles it
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