SOLUTION: 2. Suppose that a ball is thrown upward with an initial velocity of 22 ft/sec. If the ball is released at a height of 5 ft., the height equation may be written as h=-16t^2+22t+5 Wh

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 2. Suppose that a ball is thrown upward with an initial velocity of 22 ft/sec. If the ball is released at a height of 5 ft., the height equation may be written as h=-16t^2+22t+5 Wh      Log On


   

Question 93461: 2. Suppose that a ball is thrown upward with an initial velocity of 22 ft/sec. If the ball is released at a height of 5 ft., the height equation may be written as h=-16t^2+22t+5 When, to the nearest thousandth of a sec., will the ball be at a height of 9ft. on the way up?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!


To find the time the ball is in the air at 9 ft, simply let h=9

9=-16t%5E2%2B22t%2B5+

0=-16t%5E2%2B22t-4+ Subtract 9 from both sides



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%2B22%2At-4=0 ( notice a=-16, b=22, and c=-4)

t+=+%28-22+%2B-+sqrt%28+%2822%29%5E2-4%2A-16%2A-4+%29%29%2F%282%2A-16%29 Plug in a=-16, b=22, and c=-4



t+=+%28-22+%2B-+sqrt%28+484-4%2A-16%2A-4+%29%29%2F%282%2A-16%29 Square 22 to get 484



t+=+%28-22+%2B-+sqrt%28+484%2B-256+%29%29%2F%282%2A-16%29 Multiply -4%2A-4%2A-16 to get -256



t+=+%28-22+%2B-+sqrt%28+228+%29%29%2F%282%2A-16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-22+%2B-+2%2Asqrt%2857%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+=+%28-22+%2B-+2%2Asqrt%2857%29%29%2F-32 Multiply 2 and -16 to get -32

So now the expression breaks down into two parts

t+=+%28-22+%2B+2%2Asqrt%2857%29%29%2F-32 or t+=+%28-22+-+2%2Asqrt%2857%29%29%2F-32


Now break up the fraction


t=-22%2F-32%2B2%2Asqrt%2857%29%2F-32 or t=-22%2F-32-2%2Asqrt%2857%29%2F-32


Simplify


t=11+%2F+16-sqrt%2857%29%2F16 or t=11+%2F+16%2Bsqrt%2857%29%2F16


So these expressions approximate to

t=0.215635347795578 or t=1.15936465220442


So our possible solutions are:
t=0.215635347795578 or t=1.15936465220442


But since we only want the height when the ball is going up, we only care about the first solution.

So our solution is t=0.215635347795578 which is 0.216 to the nearest thousandth