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August 1, 2014

August 1, 2014

Posted by **Abbey** on Friday, April 16, 2010 at 3:33pm.

a) Find the equation that describes the motion as a function of time.

My answer was x = 0 and y=-16t^2+50t+6

b) How long is the ball in the air?

This is the part I need help on. Can I put the y equation into my graphing calculator to solve for t?

c) Determine when the ball is at maximum height. Find the maximum height.

- Math(Please help) -
**Damon**, Friday, April 16, 2010 at 3:51pmwhen is y = 0 (hits ground)?

0 = 16 t^2 -50 t - 6 (I multiplied both sides by -1)

0 = 8 t^2 -25 t -3

solve quadratic for t

t = [ 25 +/- sqrt (625 + 96) ] / 32

t = [25 +/- 26.8514]/16

use the + root the - root is before you threw it

t = 3.24 seconds

max height when velocity = 0

V = Vo -32 t

32 t = 50

t = 1.56 seconds

height = 6 + 50 (1.56) - 16 (1.56^2)

- Math(Please help) -
**Abbey**, Friday, April 16, 2010 at 3:53pmThank You!!

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