SOLUTION: My problem is: An arrow is shot from the top of a 144 foot tower with an initial velocity of 128 feet per second. The height of the arrow, h feet, after t seconds is given by the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: My problem is: An arrow is shot from the top of a 144 foot tower with an initial velocity of 128 feet per second. The height of the arrow, h feet, after t seconds is given by the       Log On


   

Question 92401: My problem is:
An arrow is shot from the top of a 144 foot tower with an initial velocity of 128 feet per second. The height of the arrow, h feet, after t seconds is given by the equation.
h=-16t^2+128t+144
a. If the arrow reaches its maximum height in 4 seconds, find its maximum height.
b. How many seconds will it take for the arrow to reach the ground?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
"a. If the arrow reaches its maximum height in 4 seconds, find its maximum height"


Lets find h when t=4

h=-16t%5E2%2B128t%2B144 Start with the given polynomial


h=-16%284%29%5E2%2B128%284%29%2B144 Plug in t=4


h=-16%2816%29%2B128%284%29%2B144 Raise 4 to the second power to get 16


h=-256%2B128%284%29%2B144 Multiply -16 by 16 to get -256


h=-256%2B512%2B144 Multiply 128 by 4 to get 512


h=400 Now combine like terms


So when t=4, h=400

So the maximum height is 400 feet



"b. How many seconds will it take for the arrow to reach the ground?"

Let h=0 and solve for t

0=-16t%5E2%2B128t%2B144



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%2B128%2At%2B144=0 ( notice a=-16, b=128, and c=144)

t+=+%28-128+%2B-+sqrt%28+%28128%29%5E2-4%2A-16%2A144+%29%29%2F%282%2A-16%29 Plug in a=-16, b=128, and c=144



t+=+%28-128+%2B-+sqrt%28+16384-4%2A-16%2A144+%29%29%2F%282%2A-16%29 Square 128 to get 16384



t+=+%28-128+%2B-+sqrt%28+16384%2B9216+%29%29%2F%282%2A-16%29 Multiply -4%2A144%2A-16 to get 9216



t+=+%28-128+%2B-+sqrt%28+25600+%29%29%2F%282%2A-16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-128+%2B-+160%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+=+%28-128+%2B-+160%29%2F-32 Multiply 2 and -16 to get -32

So now the expression breaks down into two parts

t+=+%28-128+%2B+160%29%2F-32 or t+=+%28-128+-+160%29%2F-32

Lets look at the first part:

x=%28-128+%2B+160%29%2F-32

t=32%2F-32 Add the terms in the numerator
t=-1 Divide

So one answer is
t=-1



Now lets look at the second part:

x=%28-128+-+160%29%2F-32

t=-288%2F-32 Subtract the terms in the numerator
t=9 Divide

So another answer is
t=9

So our possible solutions are:
t=-1 or t=9

However, a negative time doesn't make sense. So our only answer is

t=9

So it takes 9 seconds to reach the ground