Que: A ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where \(h=\ -5\left(t-20\right)^2+180\). What is the height of the ball once it reached its maximum height and then descended for 5 seconds?
A) 55 feet
B) 105 feet
C) 190 feet
D) 200 feet
E) 255 feet
Que: A ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where.....
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- Max@Math Revolution
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Que: A ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where.....
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Solution: We know that \(h=-5\left(t - 20\right)^2+180\).
We will first find the value for ‘t’ for which ‘h’ will be maximum.
For ‘h’ to be maximum, \(-5\left(t-20\right)^2\) should be maximum. Since \(-5\left(t-20\right)^2\) is a perfect square, therefore, \(-5\left(t-20\right)^2\) ≥ 0.
But, \(-5\left(t-20\right)^2\) will be ≤ 0 [By the property of reverse inequality]
So, for ‘h’ to be maximum \(-5\left(t-20\right)^2\) = 0
=> \(-5\left(t-20\right)^2\)
=> \(\left(t-20\right)^2\)
=> (t − 20) = 0
=> t = 20.
‘5’ seconds after ball has reached maximum height ‘h’ at t = 20 + 5 = 25.
=> \(h=-5\left(25-20\right)^2+180\)
=> \(h=-5\left(5\right)^2+180\)
=> h = −5 * 25 + 180
=> h = -125 + 180
=> h = 55 feet
A is the correct answer.
Answer A
We will first find the value for ‘t’ for which ‘h’ will be maximum.
For ‘h’ to be maximum, \(-5\left(t-20\right)^2\) should be maximum. Since \(-5\left(t-20\right)^2\) is a perfect square, therefore, \(-5\left(t-20\right)^2\) ≥ 0.
But, \(-5\left(t-20\right)^2\) will be ≤ 0 [By the property of reverse inequality]
So, for ‘h’ to be maximum \(-5\left(t-20\right)^2\) = 0
=> \(-5\left(t-20\right)^2\)
=> \(\left(t-20\right)^2\)
=> (t − 20) = 0
=> t = 20.
‘5’ seconds after ball has reached maximum height ‘h’ at t = 20 + 5 = 25.
=> \(h=-5\left(25-20\right)^2+180\)
=> \(h=-5\left(5\right)^2+180\)
=> h = −5 * 25 + 180
=> h = -125 + 180
=> h = 55 feet
A is the correct answer.
Answer A
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