To find out how far horizontally from its starting point the rocket will land, we need to determine the x-coordinate when the rocket hits the ground. This occurs when y = 0 (the height is zero).
Therefore, we need to solve the equation -0.02x^2 + 2.3x + 6 = 0 for x.
-0.02x^2 + 2.3x + 6 = 0
Multiplying by 100 to clear the decimals, we get:
-2x^2 + 230x + 600 = 0
Now, we can use the quadratic formula to find the solutions for x:
x = (-B ± sqrt(B^2 - 4AC))/(2A)
Where A = -2, B = 230, C = 600
x = (-230 ± sqrt(230^2 - 4(-2)(600)))/(2*(-2))
x = (-230 ± sqrt(52900))/(-4)
x = (-230 ± 230)/(-4)
x1 = (0)/(-4) = 0 (the starting point)
x2 = (460)/(-4) = -115
Therefore, the rocket will land 115 meters horizontally from its starting point.
A model rocket is launched from a root into a large field. The path of the rocket can be modeled by the equation y = –0.02x2 + 2.3x + 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?%0D%0A(1 point)
1 answer