A physics student stands at the top of a hill that has an elevation of 56 meters. he throws a rock and it goes up into the air and then falls back past him and lands on the ground below. The path of the rock can be modeled by the equation y=0.04x^2+1.3x+56 where x is the horizontal distance, in meters, from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground.
How far horizontally from its starting point will the rock land? round your answer to the nearest hundredth
56.00
24.54
57.04
57.26* my answer
To find the horizontal distance at which the rock will land, we need to find the value of x when y = 0. Since the equation given is y = 0.04x^2 + 1.3x + 56, we can set y equal to zero and solve for x.
0 = 0.04x^2 + 1.3x + 56
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 0.04, b = 1.3, and c = 56. Plugging these values into the quadratic formula:
x = (-1.3 ± √(1.3^2 - 4 * 0.04 * 56)) / (2 * 0.04)
Simplifying further:
x = (-1.3 ± √(1.69 - 8.96)) / 0.08
x = (-1.3 ± √(-7.27)) / 0.08
Since the value under the square root is negative, there are no real solutions to this equation. Therefore, the rock will not land on the ground, and the equation given does not accurately model the path of the rock.
To find the horizontal distance at which the rock will land, we need to determine the value of x when y is equal to 0. In other words, we need to find the x-intercept of the equation y = 0.04x^2 + 1.3x + 56.
To solve for x, we set y to 0 and solve the quadratic equation:
0 = 0.04x^2 + 1.3x + 56.
This equation does not have integer solutions, so we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a.
In this case, a = 0.04, b = 1.3, and c = 56. Plugging in these values, we have:
x = (-1.3 ± √(1.3^2 - 4*0.04*56)) / (2*0.04).
Simplifying further, we get:
x = (-1.3 ± √(1.69 - 8.96)) / 0.08.
x = (-1.3 ± √(-7.27)) / 0.08.
Since we cannot take the square root of a negative number within the realm of real numbers, we conclude that there are no real solutions for x. This means that the rock will not land on the ground below.
Therefore, none of the answer choices provided (56.00, 24.54, 57.04, 57.26) is correct.
can someone help?
you must have a typo, since all your terms are positive.
Your rock will continue to rise higher and higher.
I was expecting the square term to be negative.
so I will assume that
-.04x^2 + 1.3x + 56 = 0 (y is zero when it hits the ground)
times -100
4x^2 - 130x - 5600 = 0
x = (130 ± √106500)/8
= 57.04 or some negative