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December 20, 2014

December 20, 2014

Posted by **Jaclyn** on Friday, September 7, 2007 at 2:06am.

- physics -
**drwls**, Friday, September 7, 2007 at 3:01amYou can do this in various ways. One would be to solve the equation for height y (above the pool) vs. time, for when time t when y = 0

y = 3 + 5 t - (g/2) t^2 = 0

where g = 9.8 m/s^2 is the acceleration of gravity.

This quadratic equation will have two roots. Take the one that is positive. (t>0)

Another way to get the answer is to compute how high the diver gets (ymax), and then add the time spend going up (from y=3m to ymax) and the time spend coming back down to y=0

The time t1 spent going up is given by

g*t1 = 5

t1 = 0.51 s

The maximum height attained is

ymax = 3 + (2.5 m/s)(0.51) = 4.28 m

The time t2 that it takes to come back down from that height is given by

ymax = (g/2) t2^2

t2 = sqrt [2*ymax/g] = 0.93 s

The total time in the air is 0.51 + 0.93 s.

- physics-drwls!!!! -
**Jaclyn**, Friday, September 7, 2007 at 3:41amI can't seem to make the quadratic equation work when i put in that equation...help!

- physics-drwls!!!! -
- physics -
**drwls**, Friday, September 7, 2007 at 5:45amI gave you two ways to do the problem. They should give the same answer.

y = -4.9 t^2 + 5t +3 = 0

t = [-5 - sqrt(25 + 4*4.9*3)]/-2(4.9)

= (5 + sqrt83.8)/9.8

= 1.44 s

That agrees with my 0.51 + 0.93 = 1.44 s answer

Note that I only took the - sign in the +/- (b^2 - 4ac) term of the quadratic equation. The other term would have given me a meaningless negative answer for t.

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