## 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 [email protected] Revolution » Mon Jun 14, 2021 10:58 pm

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## Global Stats

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 [email protected] 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