A baseball player hits a baseball into the outfield. The equation h= -0.005x^2 +x+3 give the path of the ball, where (h) is the height and (x) is the horizontal distance the ball travels.
An outfielder catches 3 ft above the ground. How far has the ball traveled horizontally when the outfielder catches it.
to do this, we substitute h = 3 to the equation and solve for x:
h = -0.005x^2 + x + 3
3 = -0.005x^2 + x + 3
0 = -0.005x^2 + x
0 = x(-0.005x + 1)
x = 0 (this is extraneous)
x = 1/0.005 = 200 ft
hope this helps~ :)
Thank you so much!!! I got the first 2 but I didn't have a clue on that
To find out how far the ball has traveled horizontally when the outfielder catches it, we need to find the value of x in the equation h = -0.005x^2 + x + 3 when h = 3.
1. Substitute h = 3 into the equation:
3 = -0.005x^2 + x + 3
2. Simplify the equation by subtracting 3 from both sides:
0 = -0.005x^2 + x
3. Rearrange the equation to form a quadratic equation:
0 = -0.005x^2 + x
4. Set the equation equal to zero to solve for x:
-0.005x^2 + x = 0
5. Factor out x from the equation:
x(-0.005x + 1) = 0
6. Set each factor equal to zero and solve for x:
x = 0 or -0.005x + 1 = 0
For x = 0, it means the ball was caught at the starting point, which doesn't make sense in this context. So we ignore this solution.
For -0.005x + 1 = 0, solve for x:
-0.005x = -1
x = -1 / -0.005
x = 200
Therefore, the ball has traveled horizontally 200 feet when the outfielder catches it.