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March 29, 2015

March 29, 2015

Posted by **Andy G** on Tuesday, January 14, 2014 at 9:15pm.

a. draw the illustration.

b. for what time interval is ball more than 96 ft above ground?

c. what's the max height of ball?

d. after how many seconds does the ball reach max height?

- Pre-Calc -
**Henry**, Wednesday, January 15, 2014 at 3:40pmb. d = 80t - 16t^2 = 96

-16t^2 + 80t - 96 = 0

Divide both sides by 16:

-t^2 + 5t - 6 = 0

Use Quadratic Formula and get:

t = 2s, and 3 s.

Interval: 2< T <3.

c. h = (V^2-Vo^2)/2g

h = (0-80^2)/64 = 100 Ft.

d. V = Vo + g*t = 0 at max ht.

t = -Vo/g = -80/-32 = 2.50 s.

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