Posted by **steph** on Thursday, November 10, 2011 at 8:19pm.

Physics: a ball is thrown vertically upward with an initial velocity of 80 feet per second. the distance (in feet) of the ball from the ground after t seconds if s=80t-16tt^2

a) draw the illustration

b) for what time interval is the ball more than 96 feet above the ground?

c) what is the maximum height of the ball?

d) after how many seconds does the ball reach the maximum height?

- pre-calc -
**Reiny**, Thursday, November 10, 2011 at 9:56pm
b)

80t - 16t^2 > 96

16t^2 - 80t + 96 < 0

t^2 - 5t + 6 < 0

(t-2)(t-3) < 0

so the ball is higher than 96 for

2 < t < 3, that is, between 2 and 3 seconds

c) s = -16(t^2 - 5t ** + 25/4 - 25/4**)

= -16(t-5/2)^2 + 100

I will leave it up to you to interpret that result.

- pre-calc -
**steph**, Thursday, November 10, 2011 at 10:09pm
another great help!

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