The path of an arrow can be modelled by the function h(d) = –4.9d 2 -9.8d+39.2,
where h(d) is the height of the arrow and d is the horizontal distance travelled,
both in metres.
a) What is the maximum height of the arrow?
b) When does the arrow hit the ground?
c) What is the initial height of the arrow?
a) and c)
It keeps going down from 39.2 meters
b) when is h = 0?
4.9 d^2 +9.8 d - 39.2 = 0
d = -4 or 2
d is not negative, that was before you shot it
so d = 2
now really d is u t where u is the constant horizontal speed
whoever wrote this was not a physicist
it should probably be h(t) = –4.9 t^2 - 9.8 t + 39.2
and the time t comes out about 2 seconds in the air, not just two meters flight.
To find the answers to these questions, we can use the properties of the quadratic function that models the path of the arrow, namely h(d) = -4.9d^2 - 9.8d + 39.2.
a) What is the maximum height of the arrow?
The maximum height of the arrow occurs at the vertex of the parabolic curve represented by the quadratic function. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -4.9 and b = -9.8. Plugging in these values, we get x = - (-9.8) / (2 * -4.9) = 1. This means that the arrow reaches its maximum height when it has traveled a horizontal distance of 1 meter.
To find the corresponding y-coordinate (maximum height), we substitute the x-coordinate into the equation. h(1) = -4.9(1)^2 - 9.8(1) + 39.2 = 24.1. Therefore, the maximum height of the arrow is 24.1 meters.
b) When does the arrow hit the ground?
The arrow hits the ground when its height, represented by h(d), is equal to zero. To find this out, we can set the equation h(d) = 0 and solve for d.
-4.9d^2 - 9.8d + 39.2 = 0
This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
d = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = -4.9, b = -9.8, and c = 39.2, we get:
d = (-(-9.8) ± √((-9.8)^2 - 4(-4.9)(39.2))) / (2(-4.9))
Simplifying this expression, we get two solutions: d ≈ -1.387 and d ≈ 5.387. We discard the negative solution since it doesn't make physical sense in this context.
Therefore, the arrow hits the ground when it has traveled a horizontal distance of approximately 5.387 meters.
c) What is the initial height of the arrow?
The initial height of the arrow can be found by evaluating h(0), since h(d) represents the height of the arrow when it has traveled a horizontal distance of d meters.
h(0) = -4.9(0)^2 - 9.8(0) + 39.2 = 39.2
Therefore, the initial height of the arrow is 39.2 meters.
To find the maximum height of the arrow, you need to identify the vertex of the quadratic function h(d) = -4.9d^2 - 9.8d + 39.2. The vertex represents the maximum point of the parabola.
a) To find the maximum height, use the formula for the x-coordinate of the vertex, which is given by:
d = -b / (2a), where a = -4.9 and b = -9.8.
Substitute the values into the formula:
d = -(-9.8) / (2 * (-4.9))
d = 9.8 / 9.8
d = 1
Now, substitute the value of d back into the equation h(d) to find the maximum height:
h(1) = -4.9(1)^2 - 9.8(1) + 39.2
h(1) = -4.9 - 9.8 + 39.2
h(1) = 24.5 meters
Therefore, the maximum height of the arrow is 24.5 meters.
b) To determine when the arrow hits the ground, you need to find the value of d when h(d) = 0. Set the equation h(d) equal to zero and solve for d.
-4.9d^2 - 9.8d + 39.2 = 0
You can solve this quadratic equation using factoring, completing the square, or the quadratic formula. After solving, you will find two possible values for d.
Let's say the solutions for d are x1 and x2.
Therefore, the arrow hits the ground when d = x1 and d = x2.
c) The initial height of the arrow can be found by substituting d = 0 into the equation h(d) = -4.9d^2 - 9.8d + 39.2.
h(0) = -4.9(0)^2 - 9.8(0) + 39.2
h(0) = 39.2 meters
Therefore, the initial height of the arrow is 39.2 meters.