A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y=-.02x^2+9.5x+5.6, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land. Round your answer to the nearest hundredth meter. Show all work.
AAAaannndd the bot gets it wrong yet again!
32.76 meters? Not much of a rocket!
No, the smaller answer is -32.76 -- the answers mean that if it had been launched from the ground, it would have been 32.76 meters farther away than the roof.
So the correct answer is 468.24
(actually, you got even that wrong -- it's 475.59)
I apologize for the mistake in my earlier response. The correct answer is indeed 475.59 meters, as the rocket will land 475.59 meters horizontally from its starting point on the roof. Thank you for bring this to my attention.
Correct solution:
-.02x^2+9.5x+5.6 = 0 , divide each term by -.02
x^2 - 475x - 280 = 0
even though the numbers are bigger, I have no decimals
Looks like the bot can't use a calculator, I get
x = -.58874 , which we will reject because of the negative
or
x = 475.59 metres
You are correct. Thank you for providing the correct solution. I apologize for my mistake in my earlier responses. The rocket will land approximately 475.59 meters horizontally from its starting point on the roof.
To find the horizontal distance at which the rocket will land, we need to determine the x-coordinate when the rocket reaches the ground. In other words, we need to find the value of x when the height of the rocket (y) is equal to zero.
The given equation for the rocket's path is:
y = -0.02x^2 + 9.5x + 5.6
Setting y equal to zero, we get:
0 = -0.02x^2 + 9.5x + 5.6
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -0.02, b = 9.5, and c = 5.6. Substituting these values into the quadratic formula, we get:
x = (-(9.5) ± √((9.5)^2 - 4(-0.02)(5.6))) / (2(-0.02))
Simplifying further:
x = (-9.5 ± √(90.25 + 0.448)) / (-0.04)
x = (-9.5 ± √90.698) / (-0.04)
Using a calculator, we find that √90.698 is approximately 9.527. Plugging this value into the equation:
x = (-9.5 ± 9.527) / (-0.04)
Now, we can solve for x by considering both the positive and negative solutions separately:
1. Positive solution:
x = (-9.5 + 9.527) / (-0.04)
x = 0.027 / -0.04
x ≈ -0.675
2. Negative solution:
x = (-9.5 - 9.527) / (-0.04)
x = -19.027 / -0.04
x ≈ 475.68
Since we are dealing with a real-world scenario, the value of x cannot be negative in this context. Therefore, the rocket will land approximately 475.68 meters horizontally from its starting point. Rounding to the nearest hundredth meter, the rocket will land about 475.68 meters away.
To find out where the rocket lands, we need to find the value of x when y (the height of the rocket) is zero (when it hits the ground).
0 = -.02x^2 + 9.5x + 5.6 (substitute y=0)
We can solve for x using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a (quadratic formula)
In this equation, a=-.02, b=9.5, and c=5.6.
x = (-9.5 ± sqrt(9.5^2 - 4(-.02)(5.6))) / 2(-.02) (substitute values)
x ≈ 468.24 or x ≈ 32.76 (simplify using a calculator)
We know that the rocket is launched from the roof, so the only possible answer is x ≈ 32.76 meters. Therefore, the rocket will land about 32.76 meters horizontally from its starting point.