Posted by **lucy** on Saturday, December 15, 2007 at 2:08pm.

If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t-16t^2. What is the maximum height attained by the ball?

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**drwls**, Saturday, December 15, 2007 at 2:33pm
That would be when the velocity is zero. That happens when the derivative

80 - 32t, is zero. Therefore t = 2.5 seconds at that time, and the maximum height is

80*2.5 - 16(2.5)^2 = 100 ft

- math -
**lucely**, Saturday, December 15, 2007 at 5:38pm
what ever

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**tchrwill**, Saturday, December 15, 2007 at 3:07pm
Alternatively:

If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t-16t^2. What is the maximum height attained by the ball?

The time to reach the maximum height and zero velocity derives from Vf = Vo - gt wherer Vf = the final velocity = 0, Vo = the initial velocity = 80fps, t = the time to zero velocity and g = the deleration due to gravity.

Therefore, 0 = 80 - 32t making t = 2.5 sec.

Then, from h = 80t-16t^2, h = 80(2.5) - 16(2.5)^2 = 200 - 100 = 100 ft.

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**lucy**, Saturday, December 15, 2007 at 4:16pm
thanks!

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