SOLUTION: h(t)= c-(d-4t)^2; At time t=0, an initial height of 6 feet. Until the ball hit the ground, its height in feet after t seconds was given by the function h (at the beginning of prob

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Question 470978: h(t)= c-(d-4t)^2; At time t=0, an initial height of 6 feet. Until the ball hit the ground, its height in feet after t seconds was given by the function h (at the beginning of problem), in which c and d are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the height in feet of the ball at t=1 ?
The answer is 70, but I do not understand how they came up with this answer....please help!
Found 2 solutions by Alan3354, bucky:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
h(t)= c-(d-4t)^2; At time t=0, an initial height of 6 feet. Until the ball hit the ground, its height in feet after t seconds was given by the function h (at the beginning of problem), in which c and d are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the height in feet of the ball at t=1 ?
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Solve for c & d
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h(t)= c-(d-4t)^2
h(0) = c - d^2 = 6
h(2.5) = c - (d - 10)^2 = 106
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c - d^2 = 6
c = d^2 + 6
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c - (d - 10)^2 = 106
c - d^2 + 20d - 100 = 106
Sub for c
d^2+6 - d^2 + 20d = 206
20d = 200
d = 10
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c = 106
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h(t) = 106 - (10 - 4t)^2
h(1) = 106 - 6^2
h(1) = 70

Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
This is an exercise in finding the constants based on several pieces of knowledge that is given in the problem. The use of c and d as constants without any explanation of what they actually are is an uncommon practice. I understand the math, but don't understand why the originator of the problem made it difficult by using an uncommon exercise. No wonder nobody has attempted this. My complaint is over. On to the problem.
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The equation you were given is:
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Let's start by expanding the right side. Do that by squaring the term in the parenthesis to get:
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Remove the parenthesis by changing the signs of all the terms within to get:
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Note that the first two terms on the right side of the equal sign are constants that do not depend on time.
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The problem says that at t = 0, the height of the ball is 6 feet. Therefore, if we set the right side equal to 6 and t = 0 we can learn something. Let's do that:
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Substitute zero as the value of t and the equation reduces to:
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Now we know that:
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Remember this. You are also told that when t = 2.5 seconds the height of the ball is 106 feet. Let's now go back to the equation for h(t) and let h(2.5) = 106 and t = 2.5. The equation:
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Then becomes:
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Remember now that we found that . So substitute 6 for to make the equation:
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Multiply out the last two terms on the right side:
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and
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Substitute these values into the equation to get:
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Solve for d by first adding 100 to both sides of the equation:
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and then subtracting 6 from both sides:
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Finally, divide both sides by 20 and you have d = 10.
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Let's not lose sight of solving the problem for the height when t = 1. Let's return to the equation:
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We now know that and
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Substitute these two values into the equation and it becomes:
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Multiply out the second term on the right side to get:
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Since we are looking at t = 1 second, substitute 1 for t and you have:
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By combining the terms on the right side you confirm the answer that you were given, namely:
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feet.
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Hope this helps you. I still don't like this problem as a learning exercise in ballistic physics.