The rocket will land when y = 0 (on the ground). So we can set the equation equal to 0 and solve for x:
0 = -0.8x^2 + 12x + 25.8
Using the quadratic formula, we get:
x = (-b ± √(b^2 - 4ac)) / 2a
where a = -0.8, b = 12, and c = 25.8
Plugging in these values, we get:
x = (-12 ± √(12^2 - 4(-0.8)(25.8))) / (2(-0.8))
x = (-12 ± 17.52) / (-1.6)
We get two solutions: x ≈ -0.48 and x ≈ 15.48. The negative value doesn't make sense in this context, so we discard it. Therefore, the rocket will land approximately 15.48 meters horizontally from its starting point on the roof. Rounded to the nearest hundredth, this would be 15.48 meters.
A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y=-0.8x^2+12x+25.8 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.
5 answers
correct solution found 3 posts above this.
Btw, the bot tried this each time and got it wrong 3 times
Btw, the bot tried this each time and got it wrong 3 times
I apologize for the earlier incorrect responses. I will ensure that I am more careful in the future. Thank you for alerting me to the mistake.
The bot also can't tell time, you made the mistakes after this, not before
My apologies, you are correct. I made the mistake in my earlier response, and I appreciate your bringing it to my attention. I will be more mindful and accurate in my future responses. Thank you.