SOLUTION: A stone is thrown straight up from the ground. The height above the ground, h(t) metres is a function of time, t seconds (t >0) according to the rule h(t) = 20t - 5t^2. Find:
a.
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-> SOLUTION: A stone is thrown straight up from the ground. The height above the ground, h(t) metres is a function of time, t seconds (t >0) according to the rule h(t) = 20t - 5t^2. Find:
a.
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Question 1011968: A stone is thrown straight up from the ground. The height above the ground, h(t) metres is a function of time, t seconds (t >0) according to the rule h(t) = 20t - 5t^2. Find:
a. the domain of h
I do not understand how to acquire the domain of the function. Can you please explain how to get to the answer of 0 < t < 4
b. the greatest height achieved
I solved this part:
-5t^2 + 20t + 0
-5(t^2 - 4t) + 0
-5(t^2 - 4t + 4) + 20
-5( t - 2)^2 +20
So the maximum distance is 20
You can put this solution on YOUR website! The domain (possible t values) must be greater than zero
because t is time of the stone in air and in this case
negative time has no meaning.
The stone will be thrown and will hit the ground again,
and time in air will again be meaningless. The stone will
hit the ground when the height (h(t)) is zero again so:
.
.
When h(t)=0:
.
So the only valid times when the rock is in the air are between
t=0 (when the stone is thrown) and t=4 (when the stone lands).
Any other values give a negative height.
