Algebra
If a ball is thrown vertically upward from a height of 56ft. above ground with an initial velocity of 40ft. per second, then the height of the ball above ground t seconds after it is thrown is given by f(t)=16t^2 + 40t +56. How many seconds will elapse after the ball is thrown before it hits the ground?
I need step by step explaination please. I know the answer is 3.5 seconds.

Well, you have a function. If f(t) = the height, when does f=0?
In other words, just solve the dang quadratic!
16t^2 + 40t +56 = 0
t = 1 and 3.5
See whether you can pick the proper one.
posted by Steve

okay, so I solved as follows:
16t+40t+56=0
(4t+4)(4t+14)
4t+4=0 and 4t+14=0
4t=4 and 4t=14
4t=4/4 and 4t= 14/4
= t=1 and t = 7/2
Now, I am lost at what to do from here.posted by Velocity per second

duh, I got it. you divide 7/2 to equal 3.5
posted by Velocity per second
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