You can put this solution on YOUR website! 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?
:
There are two forces at work on the ball,
Force of gravity pulling downward: -16t^2
Initial speed upward: + 80t
:
h = -16t^2 + 80t
:
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:
:
You can see that the axis of symmetry is 2.5 sec
The max height occurs at the vertex:
:
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
:
:
Did this help
You can put this solution on YOUR website! 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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EQUATION:
h(t) = -16t^2+vot+so
Your Problem:
so=0 because it is thrown upward from the ground.
You want h(t) to be 0 because you want to know when it hits the ground.
-16t^2+80t=0
-16t(t-5)=0
t=0 or t=5
This means the ball was on the ground when time was zero and it
will be on the ground again when time is 5 seconds.
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Cheers,
Stan H.
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