SOLUTION: A BATTER HITS A PITCHED BASEBALL WHEN IT IS 3 FEET OFF THE GROUND. AFTER IT IS HIT THE HEIGHT (IN FEET) OF THE BALL IS MODELED BY H=-16T^2+80T+3 WHERE T IS THE TIME ( IN SECONDS).

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A BATTER HITS A PITCHED BASEBALL WHEN IT IS 3 FEET OFF THE GROUND. AFTER IT IS HIT THE HEIGHT (IN FEET) OF THE BALL IS MODELED BY H=-16T^2+80T+3 WHERE T IS THE TIME ( IN SECONDS).      Log On


   

Question 119166: A BATTER HITS A PITCHED BASEBALL WHEN IT IS 3 FEET OFF THE GROUND. AFTER IT IS HIT THE HEIGHT (IN FEET) OF THE BALL IS MODELED BY H=-16T^2+80T+3 WHERE T IS THE TIME ( IN SECONDS). HOW LONG WILL IT TAKE THE BALL TO HIT THE GROUND? ROUND TO THE NEAREST HUNDEREDTH. H= 0. MAKE SURE THIS EQUATION IS IN STANDARD FORM.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The ball will hit the ground when h=0


h=-16t%5E2%2B80t%2B3 Start with the given equation


0=-16t%5E2%2B80t%2B3 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%2B3=0 ( notice a=-16, b=80, and c=3)




t+=+%28-80+%2B-+sqrt%28+%2880%29%5E2-4%2A-16%2A3+%29%29%2F%282%2A-16%29 Plug in a=-16, b=80, and c=3



t+=+%28-80+%2B-+sqrt%28+6400-4%2A-16%2A3+%29%29%2F%282%2A-16%29 Square 80 to get 6400



t+=+%28-80+%2B-+sqrt%28+6400%2B192+%29%29%2F%282%2A-16%29 Multiply -4%2A3%2A-16 to get 192



t+=+%28-80+%2B-+sqrt%28+6592+%29%29%2F%282%2A-16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-80+%2B-+8%2Asqrt%28103%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%28103%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%28103%29%29%2F-32 or t+=+%28-80+-+8%2Asqrt%28103%29%29%2F-32


Now break up the fraction


t=-80%2F-32%2B8%2Asqrt%28103%29%2F-32 or t=-80%2F-32-8%2Asqrt%28103%29%2F-32


Simplify


t=5+%2F+2-sqrt%28103%29%2F4 or t=5+%2F+2%2Bsqrt%28103%29%2F4


So these expressions approximate to

t=-0.0372228912730548 or t=5.03722289127306


So our possible solutions are:
t=-0.0372228912730548 or t=5.03722289127306


However, since a negative time doesn't make sense, our only solution is t=5.03722289127306


So it takes about 5.03 seconds for the ball to hit the ground.