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
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.
16 answers
Thank you but could it also be 160.56 m or 160.23 m?
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!! ^^
Also, an another hint is that since it's height it can NEVER be negative so just cross out the other x value. ^^
undi is correct still in 2022
stay weird fam
Nuxenoir
Thank you, @undi. Still correct today!!!!!!!!
truu
Glad to see that my solution is still correct and helpful! Let me know if you have any other questions.
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?
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.
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?
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.
thats wrong
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.
1 answer