Posted by **Amy~** on Tuesday, September 7, 2010 at 10:58pm.

A ball is thrown upward from the top of a building which is 96ft tall w/ Initial velocity of 80 ft per sec.

Distance s(ft) of ball from the ground after t sec is s=96+80t-16t^(2)

1) how many sec. will the ball pass the top of the building when it is coming down?

How would I solve this?

- Math: Word Problem -
**bobpursley**, Tuesday, September 7, 2010 at 11:11pm
set s(t) to 96 and solve.

96=96+80t-16t^2

solve for t

- Math: Word Problem -
**Amy~**, Tuesday, September 7, 2010 at 11:19pm
why would I set 96 for s(t) because that's the height of the building ?

but it's when the ball passes the top of the building

Also I tried solving it I got to

-16t(t-5)= 0 what should I do next

- Math: Word Problem -
**Amy~**, Tuesday, September 7, 2010 at 11:35pm
I finished solving it. thank you

- Math: Word Problem -
**katirina**, Wednesday, September 8, 2010 at 12:28am
riddle: my ones number is double my tens number, I'am less than 70 and more than 60

- Math: Word Problem -
**Suman**, Friday, April 24, 2015 at 12:19pm
A ball is thrown vertically upward with velocity 100ft/sec and height of building is 40ft what is maximum height reached by ball and what is velocity when ball hits the ground

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