SOLUTION: Hi, I could use some help with this problem. I had thought there was a similar one in here, but I couldn't find it again. Thank you! An object is thrown down from the top of a

Question 71240This question is from textbook College Algebra Essentials
: Hi, I could use some help with this problem. I had thought there was a similar one in here, but I couldn't find it again.
Thank you!
An object is thrown down from the top of a building 1280 ft tall with an initial velocity of 32 ft per second. The distance s(in feet) of the object from the ground after t seconds is s=1280-32t-16t^2
a. When will the object strike the ground?
b. What is the height of the object after 4 seconds?
Thank you. This question is from textbook College Algebra Essentials

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
In this problem the equation that you are given is the only thing you need to work with. It already accounts for the 1280 ft height of the building and the initial velocity that the object is launched with. And it accounts for the effect of gravity.
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As to when will the object hit the ground ... At the instant it hits the ground the height above the ground will be zero. So substitute 0 for "s" in the equation to get:
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0 = 1280 - 32t - 16t^2 and solve for t. Notice that each term on the right is divisible by -16. Therefore, if you divide both sides of the equation by -16 the resulting equation is:
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0 = -80 + 2t + t^2
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Let's just transpose the sides of the equation and arrange the terms in descending powers of t to get:
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t^2 + 2t - 80 = 0
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This equation can be factored into:
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(t + 10)* (t - 8) = 0
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And the equation will be true if either of the two factors is equal to zero. So by setting each factor equal to zero we get:
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t + 10 = 0 and t - 8 = 0
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If we solve these two equations we find t has two possibilities ... it either equals -10 or it equals +8. It makes no sense that t = -10 since the object didn't even start to fall until t = 0. So we can discard that answer. The result is that t = +8 is the only valid answer. The object hits the ground (reaches 0 height) at 8 seconds after it is launched downward.
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And the second problem asks what the height of the object is after 4 seconds. We can tell from the first answer that the object is still falling after 4 seconds because it hasn't yet reached the ground -- which it will do after falling 8 seconds.
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So just return to the equation:
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s = 1280 - 32t - 16t^2
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and substitute 4 in place of t. The equation becomes:
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s = 1280 - 32*4 - 16*4^2
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Do the math. 32*4 = 128 and 16*4^2 = 16*16 = 256. So the equation now is:
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s = 1280 - 128 - 256 and the result is that s = 896 feet. The object is 896 feet above the ground after it has been falling for 4 seconds.
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Hope this helps to give you some insight into the type of equation that is used to work falling body problems. Note how the initial height fits into this equation. Notice also where the initial downward velocity fits in the equation. Its sign is minus because it is downward. If you had initially thrown it upward at 32 feet per second it would have a plus sign. And the final term (the -16t^2) is the term that accounts for the effects of gravity. It has a minus sign because it is pulling the object downward and is subtracting from the initial height.

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