SOLUTION: A baseball player has hit a ball. The height, y, of the ball x seconds after it is hit is given by y= -16x^2 + 80x + 3. At what time will the ball reach its maximum height? I used

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A baseball player has hit a ball. The height, y, of the ball x seconds after it is hit is given by y= -16x^2 + 80x + 3. At what time will the ball reach its maximum height? I used      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 195718: A baseball player has hit a ball. The height, y, of the ball x seconds after it is hit is given by y= -16x^2 + 80x + 3.
At what time will the ball reach its maximum height? I used the quadratic formula and my answer was 2.04
What is the maximum height of the ball? I used the original formula.My answer was 232.78
When will the ball hit the ground? My answer was 4.08
Is this right??
thanks
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
At what time will the ball reach its maximum height?
What is the maximum height of the ball?

First, we need the vertex. The vertex will tell us what the max height will be:


In order to find the vertex, we first need to find the x-coordinate of the vertex.


To find the x-coordinate of the vertex, use this formula: x=%28-b%29%2F%282a%29.


x=%28-b%29%2F%282a%29 Start with the given formula.


From y=-16x%5E2%2B80x%2B3, we can see that a=-16, b=80, and c=3.


x=%28-%2880%29%29%2F%282%28-16%29%29 Plug in a=-16 and b=80.


x=%28-80%29%2F%28-32%29 Multiply 2 and -16 to get -32.


x=5%2F2 Reduce.


So the x-coordinate of the vertex is x=5%2F2. Note: this means that the axis of symmetry is also x=5%2F2.


Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.


y=-16x%5E2%2B80x%2B3 Start with the given equation.


y=-16%285%2F2%29%5E2%2B80%285%2F2%29%2B3 Plug in x=5%2F2.


y=-16%2825%2F4%29%2B80%285%2F2%29%2B3 Square 5%2F2 to get 25%2F4.


y=-100%2B80%285%2F2%29%2B3 Multiply -16 and 25%2F4 to get -100.


y=-100%2B200%2B3 Multiply 80 and 5%2F2 to get 200.


y=103 Combine like terms.


So the y-coordinate of the vertex is y=103.


So the vertex is .


which in decimal form is .


So this means that at 2.5 seconds, the ball will be at the max height of 103.





When will the ball hit the ground?


y=+-16x%5E2+%2B+80x+%2B+3 Start with the given equation.


0=+-16x%5E2+%2B+80x+%2B+3 Plug in y=0


Notice we have a quadratic equation in the form of 0=ax%5E2%2Bbx%2Bc where a=-16, b=80, and c=3


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%2880%29+%2B-+sqrt%28+%2880%29%5E2-4%28-16%29%283%29+%29%29%2F%282%28-16%29%29 Plug in a=-16, b=80, and c=3


x+=+%28-80+%2B-+sqrt%28+6400-4%28-16%29%283%29+%29%29%2F%282%28-16%29%29 Square 80 to get 6400.


x+=+%28-80+%2B-+sqrt%28+6400--192+%29%29%2F%282%28-16%29%29 Multiply 4%28-16%29%283%29 to get -192


x+=+%28-80+%2B-+sqrt%28+6400%2B192+%29%29%2F%282%28-16%29%29 Rewrite sqrt%286400--192%29 as sqrt%286400%2B192%29


x+=+%28-80+%2B-+sqrt%28+6592+%29%29%2F%282%28-16%29%29 Add 6400 to 192 to get 6592


x+=+%28-80+%2B-+sqrt%28+6592+%29%29%2F%28-32%29 Multiply 2 and -16 to get -32.


x+=+%28-80+%2B-+8%2Asqrt%28103%29%29%2F%28-32%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


x+=+%28-80%2B8%2Asqrt%28103%29%29%2F%28-32%29 or x+=+%28-80-8%2Asqrt%28103%29%29%2F%28-32%29 Break up the expression.


So the possible answers are x+=+%28-80%2B8%2Asqrt%28103%29%29%2F%28-32%29 or x+=+%28-80-8%2Asqrt%28103%29%29%2F%28-32%29


which approximate to x=-0.037 or x=5.037


Since a negative time doesn't make sense, this means that the only solution is x=5.037


So it will take about 5.037 seconds for the ball to hit the ground.


Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A baseball player has hit a ball. The height, y, of the ball x seconds after it is hit is given by y= -16x^2 + 80x + 3.
At what time will the ball reach its maximum height? I used the quadratic formula and my answer was 2.04
------------------------------
Find the apogee, which is the vertex of the parabola:
hmax = -b/2a
= -80/(-32) = 2.5 seconds
---------------------------
What is the maximum height of the ball? I used the original formula.My answer was 232.78
It's at time = 2.5 seconds.
y = -16*2.5^2 + 80*2.5 + 3
ymax = 103 feet
-----------------
When will the ball hit the ground? My answer was 4.08
Set y = 0
-16x^2 + 80x + 3 = 0
x = 2.5 + (sqrt(103))/4 seconds
x = ~ 5.037222891 seconds
It's not 5 seconds even, because the ball started up from 3' elevation.