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:
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.
duh, I got it. you divide 7/2 to equal 3.5