SOLUTION: I am halfway through this problem and have been working on it/been stuck on it for quite some time. I think I got part a correct, but part b has stumped me. I've tried factoring it

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am halfway through this problem and have been working on it/been stuck on it for quite some time. I think I got part a correct, but part b has stumped me. I've tried factoring it      Log On


   

Question 995100: I am halfway through this problem and have been working on it/been stuck on it for quite some time. I think I got part a correct, but part b has stumped me. I've tried factoring it, but I get stuck. I figured that I probably need to do the quadratic formula to solve it, but I cant seem to figure it out with the instructions in my book.
The question is: If an archer shoots an arrow straight upward with an initial velocity of 160 ft/sec from a height of 8ft, then its height above ground in feet at time t in seconds is given by the function h(t) = -16t^2 + 160t + 8
a. What is the maximum height reached by the arrow?
I got 5 seconds. [Correct me if I'm wrong.]
0 = -32t + 160
-160 = -32t
-160/-32 = t
5 = t
b. How long does it take for the arrow to reach the ground?
So i get it down to:
0 = -16(t^2 + 10t -1/2)
...but don't know how to go from there without screwing it up.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
For what t is h(t) maximum? This happens at the symmetry axis, which will be vertex of h(t).

Find the roots and determine the value of t at the midpoint of the roots.

-16t%5E2%2B160t%2B8
-8%282t%5E2%2B20t%2B1%29
-8(2t__1)(t__1)-----if you want to TRY to factorize.
but ignoring this possibility,
roots %28-20%2Bsqrt%28400-4%2A2%29%29%2F%282%2A2%29
%28-20%2B2sqrt%28100-2%29%29%2F4
%28-20%2B2sqrt%2898%29%29%2F4
-10%2Bsqrt%2898%29%2F2
-10%2Bsqrt%282%2A49%29%2F2
highlight%28-10%2B7sqrt%282%29%2F2%29

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
I am halfway through this problem and have been working on it/been stuck on it for quite some time. I think I got part a correct, but part b has stumped me. I've tried factoring it, but I get stuck. I figured that I probably need to do the quadratic formula to solve it, but I cant seem to figure it out with the instructions in my book.
The question is: If an archer shoots an arrow straight upward with an initial velocity of 160 ft/sec from a height of 8ft, then its height above ground in feet at time t in seconds is given by the function h(t) = -16t^2 + 160t + 8
a. What is the maximum height reached by the arrow?
I got 5 seconds. [Correct me if I'm wrong.]
0 = -32t + 160
-160 = -32t
-160/-32 = t
5 = t
b. How long does it take for the arrow to reach the ground?
So i get it down to:
0 = -16(t^2 + 10t -1/2)
...but don't know how to go from there without screwing it up.
a.
h%28t%29+=+-+16t%5E2+%2B+160t+%2B+8
Your answer for "a" cannot be correct. You have 5 seconds but it asks for the MAXIMUM HEIGHT reached. 
The 5 seconds you got is actually the time at which the MAXIMUM HEIGHT occurs. Therefore, substitute 5 for "t" in the equation: h%28t%29+=+-+16t%5E2+%2B+160t+%2B+8 to get: h%285%29+=+-+16%285%29%5E2+%2B+160%285%29+%2B+8,
or h%285%29+=+408. Thus, the MAXIMUM HEIGHT of highlight_green%28408%29 ft was reached at a time of 5 seconds after the arrow was shot.

b.
To find the time it hits the ground, you are correct in setting the equation equal to 0: 0+=+-+16t%5E2+%2B+160t+%2B+8. Since this cannot be factored it can be reduced to 0+=+2t%5E2+-+20t+-+1,

which can now be solved using the quadratic equation formula. The negative root/value should be IGNORED, but the positive value (should be: highlight_green%28t+=+10.0498%29 seconds) should be accepted as 

the time the arrow hits the ground.