Question 118276
A ball is thrown directly upward from ground level with an initial speed of 80 ft/s. How high will it go and when will it return to the ground?
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There are two forces at work on the ball, 
Force of gravity pulling downward: -16t^2
Initial speed upward: + 80t
:
h = -16t^2 + 80t
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When the ball hits the ground; h= 0
-16t^2 + 80t = 0
:
Simplify and change the signs, factor out -16t
-16t(t - 5) = 0
Two solutions
-16t = 0
t = 0; when the ball is thrown from ground level
and
t = +5 seconds elapse when the ball hits the ground
:
A graph of this shows time in seconds as x and height in feet as y:
{{{ graph( 300, 200, -4, 8, -20, 120, -16x^2+80x) }}}
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You can see that the axis of symmetry is 2.5 sec 
The max height occurs at the vertex:
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Substitute 2.5 in the original equation
h = -16(2.5^2) + 80(2.5)
h = -16(6.25) + 200
h = -100 + 200
h = +100 ft is the max height
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