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A ball, t seconds after it was thrown up, is at a height of h feet, wh
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19 Feb 2024, 10:01
19 Feb 2024, 10:01
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10% (01:59) wrong based on 10 sessions
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A ball, t seconds after it was thrown up, is at a height of h feet, where \(h = -2(t -5)^2 + 100\). What is the height of the ball now once it reached its maximum height and then descended for 5 seconds?
(A) 36 feet
(B) 50 feet
(C) 64 feet
(D) 72 feet
(E) 96 feet
(A) 36 feet
(B) 50 feet
(C) 64 feet
(D) 72 feet
(E) 96 feet
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Re: A ball, t seconds after it was thrown up, is at a height of h feet, wh
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19 Feb 2024, 14:18
19 Feb 2024, 14:18
2
We can see that max height is reached when t = 5, because that makes 2(t-5)^2 = 0.
To find height after 5 more seconds, set t = 5+5 = 10.
\(h=−2(10−5)^2+100=100-50=50\)
To find height after 5 more seconds, set t = 5+5 = 10.
\(h=−2(10−5)^2+100=100-50=50\)
gmatclubot
Re: A ball, t seconds after it was thrown up, is at a height of h feet, wh [#permalink]
19 Feb 2024, 14:18
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