SOLUTION: If an object is projected vertically upward from ground level with an initial velocity of 280 ft/sec, then its distance s above the ground after t seconds is given by s = &#8722;

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Question 950715: If an object is projected vertically upward from ground level with an initial velocity of 280 ft/sec, then its distance s above the ground after t seconds is given by
s = −14t2 + 280t.
For what values of t will the object be more than 1050 feet above the ground?

______ < t < ______
Found 2 solutions by MathLover1, josmiceli:
Answer by MathLover1(20849) (Show Source):
You can put this solution on YOUR website!

For what values of will the object be more than feet above the ground?

..............simplify, both sides divide by
......factor completely

solutions:
=>
=>
combined:

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!

----------------------
Set ft

I notice that , so I
can say:

and

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ft for
sec
sec
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sec
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Here's the plot:

SOLUTION: If an object is projected vertically upward from ground level with an initial velocity of 280 ft/sec, then its distance s above the ground after t seconds is given by s = &#8722;

SOLUTION: If an object is projected vertically upward from ground level with an initial velocity of 280 ft/sec, then its distance s above the ground after t seconds is given by s = −