SOLUTION: Solve. If a ball thrown vertically upward from a height of 5ft, with an initial velocity of 80 ft/s, it's height(h) after (t) second is given by h=-16t squared + 80t + 5. ow lo

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Solve. If a ball thrown vertically upward from a height of 5ft, with an initial velocity of 80 ft/s, it's height(h) after (t) second is given by h=-16t squared + 80t + 5. ow lo      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 122762: Solve.
If a ball thrown vertically upward from a height of 5ft, with an initial velocity of 80 ft/s, it's height(h) after (t) second is given by h=-16t squared + 80t + 5. ow long does it take the ball to return to the ground?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
When the ball hits the ground, this means h=0 (in other words, the height is zero)

h+=+-16t%5E2+%2B+80t+%2B+5 Start with the given equation


0=+-16t%5E2+%2B+80t+%2B+5 Plug in h=0.


Let's use the quadratic formula to solve for t:


Starting with the general quadratic

at%5E2%2Bbt%2Bc=0

the general solution using the quadratic equation is:

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve -16%2At%5E2%2B80%2At%2B5=0 ( notice a=-16, b=80, and c=5)




t+=+%28-80+%2B-+sqrt%28+%2880%29%5E2-4%2A-16%2A5+%29%29%2F%282%2A-16%29 Plug in a=-16, b=80, and c=5



t+=+%28-80+%2B-+sqrt%28+6400-4%2A-16%2A5+%29%29%2F%282%2A-16%29 Square 80 to get 6400



t+=+%28-80+%2B-+sqrt%28+6400%2B320+%29%29%2F%282%2A-16%29 Multiply -4%2A5%2A-16 to get 320



t+=+%28-80+%2B-+sqrt%28+6720+%29%29%2F%282%2A-16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-80+%2B-+8%2Asqrt%28105%29%29%2F%282%2A-16%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



t+=+%28-80+%2B-+8%2Asqrt%28105%29%29%2F-32 Multiply 2 and -16 to get -32

So now the expression breaks down into two parts

t+=+%28-80+%2B+8%2Asqrt%28105%29%29%2F-32 or t+=+%28-80+-+8%2Asqrt%28105%29%29%2F-32


Now break up the fraction


t=-80%2F-32%2B8%2Asqrt%28105%29%2F-32 or t=-80%2F-32-8%2Asqrt%28105%29%2F-32


Simplify


t=5+%2F+2-sqrt%28105%29%2F4 or t=5+%2F+2%2Bsqrt%28105%29%2F4


So these expressions approximate to

t=-0.0617376914898995 or t=5.0617376914899


So our possible solutions are:
t=-0.0617376914898995 or t=5.0617376914899





However, since a negative time doesn't make sense, our only solution is

t=5.0617376914899



-----------------------------

Answer:


So at around 5.06 seconds, the ball will hit the ground.