SOLUTION: The equation h=-16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground .

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: The equation h=-16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground .       Log On


   

Question 121755: The equation h=-16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground . Find the time it takes for the arrow to reach a height of 180 ft?
t = 180/112
t = 1.6 seconds
Not sure if I'm doing the problem right but just double checking.
Found 2 solutions by jim_thompson5910, algebrapro18:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
h+=+-16t%5E2+%2B+112t Start with the given equation


180+=+-16t%5E2+%2B+112t Plug in h=180



16t%5E2-112t%2B180=0 Get everything to the left side

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%5E2-112%2At%2B180=0 ( notice a=16, b=-112, and c=180)




t+=+%28--112+%2B-+sqrt%28+%28-112%29%5E2-4%2A16%2A180+%29%29%2F%282%2A16%29 Plug in a=16, b=-112, and c=180



t+=+%28112+%2B-+sqrt%28+%28-112%29%5E2-4%2A16%2A180+%29%29%2F%282%2A16%29 Negate -112 to get 112



t+=+%28112+%2B-+sqrt%28+12544-4%2A16%2A180+%29%29%2F%282%2A16%29 Square -112 to get 12544 (note: remember when you square -112, you must square the negative as well. This is because %28-112%29%5E2=-112%2A-112=12544.)



t+=+%28112+%2B-+sqrt%28+12544%2B-11520+%29%29%2F%282%2A16%29 Multiply -4%2A180%2A16 to get -11520



t+=+%28112+%2B-+sqrt%28+1024+%29%29%2F%282%2A16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28112+%2B-+32%29%2F%282%2A16%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



t+=+%28112+%2B-+32%29%2F32 Multiply 2 and 16 to get 32

So now the expression breaks down into two parts

t+=+%28112+%2B+32%29%2F32 or t+=+%28112+-+32%29%2F32

Lets look at the first part:

x=%28112+%2B+32%29%2F32

t=144%2F32 Add the terms in the numerator
t=9%2F2 Divide

So one answer is
t=9%2F2



Now lets look at the second part:

x=%28112+-+32%29%2F32

t=80%2F32 Subtract the terms in the numerator
t=5%2F2 Divide

So another answer is
t=5%2F2

So our solutions are:
t=9%2F2 or t=5%2F2 (which are t=4.5 or t=2.5 respectively in decimal form)



So it takes 2.5 seconds for the arrow to climb to 180 ft and then 4.5 seconds for the arrow to fall back to 180 ft

Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
Well that was a nice try but wrong. What you need to do is solve the equation:
180 = -16t^2+112t for t and see what you get.
so solving we get:

180 = -16t^2+112t --> Subtract 180 from both sides
0 = -16t^2+112t-180 --> factor out a 4 from the right hand side
0 = 4(-4t^2+18t-45) --> divide both sides by 4
0 = -4t^2+18t-45 --> factor
0 = (-2t+5)(2t-9) --> set each factor equal to zero and solve
-2t + 5 = 0 --> subtract 5 from both sides
-2t = -5 --> divide both sides by -2
t = 5/2 or 2.5
2t - 9 = 0 --> Add 9 to both sides
2t = 9 --> Divide both sides by 2
t = 9/2 or 4.5
so the arrow will reach a hight of 180 feet at 2.5 seconds and 4.5 seconds.