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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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Que: A ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where.....

by Max@Math Revolution » Mon Jun 14, 2021 10:58 pm

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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 \(h=\ -5\left(t-20\right)^2+180\). 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
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Re: Que: A ball thrown up in the air is at a height of h feet, t seconds after it was thrown, where.....

by Max@Math Revolution » Wed Jun 16, 2021 7:54 pm
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
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