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December 18, 2014

December 18, 2014

Posted by **lily** on Thursday, April 7, 2011 at 11:52pm.

- MATH URGENT! -
**Jai**, Friday, April 8, 2011 at 12:14amfirst we observe that the motion of the ball is uniformly accelerated motion (because of acceleration due to gravity while it's in the air), thus we use this formula to find time:

h = ho + vo*t - (1/2)*g*t^2

where

ho = initial height

vo = initial velocity

t = time

g = acceleration due to gravity (9.8 m/s^2 = 32 ft/s^2)

making the ground as the reference or the origin (thus at this point, the height = h = 0), ho is thus equal to 350

substituting,

0 = 350 + 20*t - 1/2*32*t^2

0 = 350 + 20t - 16t^2

since it's a quadratic equation, we either factor it or use quadratic formula,

t = [-b +- sqrt(b^2 - 4ac)]/(2a)

subsituting,

t = [-20 +- sqrt(20^2 - 4(-16)(350))]/(2(-16))

t = [-20 +- 150.997]/(-32)

t = -4.09

t = 5.34

since time is always positive, we get the positive root:

t = 5.34 s

hope this helps~ :)

- MATH URGENT! -
**MattsRiceBowl**, Friday, April 8, 2011 at 12:14amIt depends on many factors not said in the problem. How high does the ball go before it starts coming down, for example.

The problem is not solvable.

- MATH URGENT! -
**MattsRiceBowl**, Friday, April 8, 2011 at 12:15amOk. Someone else was smarter than I. I should have waited, I guess.

- MATH URGENT! -
**lily**, Friday, April 8, 2011 at 12:42amthanks

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