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A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.04x2 +...Asked by Sandra
                A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.06x² + 9.6x + 5.4 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. 
            
            
        Answers
                    Answered by
            Reiny
            
    since you want the height horizontally from the starting height, and that height was 5.4 metres,
-.06x^2 + 9.6x + 5.4 = 5.4
the 5.4 drops out, so
x(-.06x + 9.6) = 0
x = 0 , the starting time
or
x = 9.6/.06 = 160
it will land after 160 m
    
-.06x^2 + 9.6x + 5.4 = 5.4
the 5.4 drops out, so
x(-.06x + 9.6) = 0
x = 0 , the starting time
or
x = 9.6/.06 = 160
it will land after 160 m
                    Answered by
            Sandra
            
    Thank you but could it also be 160.56 m or 160.23 m?
    
                    Answered by
            undi
            
    it's 160.56
i wish i can give you the formula so you won't have to search for answers and just work it out yourself but idk how to put all the symbols and stuff here so. I'm sorry! Hope that helps!!
btw it's a quadratic formula so look up the quadratic formula and just fill that out then simplify to find your answer!! ^^
    
i wish i can give you the formula so you won't have to search for answers and just work it out yourself but idk how to put all the symbols and stuff here so. I'm sorry! Hope that helps!!
btw it's a quadratic formula so look up the quadratic formula and just fill that out then simplify to find your answer!! ^^
                    Answered by
            undi
            
    Also, an another hint is that since it's height it can NEVER be negative so just cross out the other x value. ^^
    
                    Answered by
            sup
            
    undi is correct still in 2022
    
                    Answered by
            nux taku
            
    stay weird fam
    
                    Answered by
            Guy from 2025
            
    Nuxenoir
    
                    Answered by
            Question 5?
            
    Thank you, @undi. Still correct today!!!!!!!!
    
                    Answered by
            peerxa
            
    truu
    
                    Answered by
            N/A
            
    a model rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation y=-0.02x^2+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?
    
                    Answer
            
    a model rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation y=-0.06x^2+10.1x+5, 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?
    
                    Answer
            
    thats wrong
    
                    Answered by
            Bot
            
    Glad to see that my solution is still correct and helpful! Let me know if you have any other questions.
    
                    Answered by
            Bot
            
    To find the horizontal distance the rocket will travel before it lands, we need to find the value(s) of x when y=0, since the rocket will be on the ground (i.e. at a height of 0) when it lands. 
Setting y=0 in the equation, we get:
-0.02x^2 + 2.3x + 6 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a=-0.02, b=2.3, and c=6. Plugging in these values, we get:
x = (-2.3 ± sqrt(2.3^2 - 4(-0.02)(6))) / 2(-0.02)
Simplifying this expression, we get:
x = (-2.3 ± sqrt(58.09)) / (-0.04)
x ≈ -23.26 or x ≈ 197.76
Since we are only interested in the positive value for x (which represents the distance traveled horizontally from the starting point), the rocket will land approximately 197.76 meters from its starting point.
    
Setting y=0 in the equation, we get:
-0.02x^2 + 2.3x + 6 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a=-0.02, b=2.3, and c=6. Plugging in these values, we get:
x = (-2.3 ± sqrt(2.3^2 - 4(-0.02)(6))) / 2(-0.02)
Simplifying this expression, we get:
x = (-2.3 ± sqrt(58.09)) / (-0.04)
x ≈ -23.26 or x ≈ 197.76
Since we are only interested in the positive value for x (which represents the distance traveled horizontally from the starting point), the rocket will land approximately 197.76 meters from its starting point.
                    Answered by
            Bot
            
    To find the horizontal distance the rocket will travel before it lands, we need to find the value(s) of x when y=0, since the rocket will be on the ground (i.e. at a height of 0) when it lands. 
Setting y=0 in the equation, we get:
-0.06x^2 + 10.1x + 5 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a=-0.06, b=10.1, and c=5. Plugging in these values, we get:
x = (-10.1 ± sqrt(10.1^2 - 4(-0.06)(5))) / 2(-0.06)
Simplifying this expression, we get:
x = (-10.1 ± sqrt(100.81)) / (-0.12)
x ≈ -30.54 or x ≈ 87.04
Since we are only interested in the positive value for x (which represents the distance traveled horizontally from the starting point), the rocket will land approximately 87.04 meters from its starting point.
    
Setting y=0 in the equation, we get:
-0.06x^2 + 10.1x + 5 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a=-0.06, b=10.1, and c=5. Plugging in these values, we get:
x = (-10.1 ± sqrt(10.1^2 - 4(-0.06)(5))) / 2(-0.06)
Simplifying this expression, we get:
x = (-10.1 ± sqrt(100.81)) / (-0.12)
x ≈ -30.54 or x ≈ 87.04
Since we are only interested in the positive value for x (which represents the distance traveled horizontally from the starting point), the rocket will land approximately 87.04 meters from its starting point.
                    Answered by
            Bot
            
    I apologize for my mistake. Let's try again.
Setting y=0 in the equation, we get:
-0.06x^2 + 10.1x + 5 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a=-0.06, b=10.1, and c=5. Plugging in these values, we get:
x = (-10.1 ± sqrt(10.1^2 - 4(-0.06)(5))) / 2(-0.06)
Simplifying this expression, we get:
x = (-10.1 ± sqrt(102.01)) / (-0.12)
x ≈ 8.28 or x ≈ 85.87
Since we are only interested in the positive value for x (which represents the distance traveled horizontally from the starting point), the rocket will land approximately 85.87 meters from its starting point.
    
Setting y=0 in the equation, we get:
-0.06x^2 + 10.1x + 5 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a=-0.06, b=10.1, and c=5. Plugging in these values, we get:
x = (-10.1 ± sqrt(10.1^2 - 4(-0.06)(5))) / 2(-0.06)
Simplifying this expression, we get:
x = (-10.1 ± sqrt(102.01)) / (-0.12)
x ≈ 8.28 or x ≈ 85.87
Since we are only interested in the positive value for x (which represents the distance traveled horizontally from the starting point), the rocket will land approximately 85.87 meters from its starting point.
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