A rock is thrown upward from a bridge that is 88 feet above a road. The rock reaches its maximum height above the road 0.67 seconds after it is thrown and contacts the road 3.18 seconds after it was thrown. Define a quadratic function, f, that gives the height of the rock above the road (in feet) in terms of the number of seconds elapsed since the rock was thrown, t
you know that if v is the initial upward velocity, v-9.8*0.67 = 0
Plug that into the usual equation
h(t) = 88 + 6.566t - 4.9t^2
Then check that h(3.18) = 0
Or, since a parabola is symmetric, you know that
h(t) = a(t-0.67)^2 + 88
h(3.18) = 0, so
a(3.18-0.67)^2 + 88 = 0
h(t) = -13.97(t-0.67)^2+88
Hmmm. The two functions don't match, but the second fits. Maybe you can check my math.
To define the quadratic function, f, that gives the height of the rock above the road in terms of the number of seconds elapsed since the rock was thrown, we need to consider the parameters of the problem.
We know that the rock reaches its maximum height 0.67 seconds after it is thrown. At the maximum height, the vertical velocity of the rock is zero.
Additionally, we have the information that the rock contacts the road 3.18 seconds after it was thrown. At this time, the height of the rock is zero.
Given these conditions, we can use the standard form of a quadratic function:
f(t) = at^2 + bt + c
where t is the time in seconds and f(t) is the height of the rock above the road.
To find the quadratic function, we need to determine the coefficients a, b, and c.
Step 1: Finding the coefficient a:
Since the vertex is the maximum point of the parabola, and the rock reaches its maximum height 0.67 seconds after it is thrown, we know that the vertex is located at (0.67, h_max).
The formula for the x-coordinate of the vertex, t = -b/(2a), gives us:
0.67 = -b / (2a)
Step 2: Finding the coefficient c:
Since the rock contacts the road 3.18 seconds after it was thrown, we know that f(3.18) = 0. This means that the height of the rock at t = 3.18 seconds is zero.
Substituting t = 3.18 into the quadratic function, we get:
0 = a*(3.18)^2 + b*(3.18) + c
Step 3: Finding the coefficient b:
We have two equations from Steps 1 and 2.
First, rearranging the equation from Step 1, we get:
b = -0.67 * 2a
Next, we substitute this value of b into the second equation from Step 2 to obtain:
0 = a*(3.18)^2 + (-0.67 * 2a)*(3.18) + c
Now we have a system of two equations (from Steps 1 and 3) with two unknowns (a and c). We can solve this system to find the coefficients a, b, and c.
Step 4: Solving the system of equations:
Using the equations from Step 1 and Step 3, we can substitute b = -0.67 * 2a into the second equation from Step 3:
0 = a*(3.18)^2 - 2*0.67*a*(3.18) + c
Multiplying out and rearranging gives us:
0 = 10.1368a - 4.2732a + c
Simplifying further:
0 = 5.8636a + c
Now we have two equations:
0.67 = -b / (2a)
0 = 5.8636a + c
Solving these equations simultaneously will give us the values of a, b, and c.
Step 5: Solving the system of equations:
From the equation 0.67 = -b / (2a), we can solve for b:
b = -0.67 * 2a
Substituting this value of b into the equation 0 = 5.8636a + c, we get:
0 = 5.8636a + c
Next, substituting the expression for b into the equation 0 = a*(3.18)^2 + (-0.67 * 2a)*(3.18) + c, we get:
0 = a*(3.18)^2 + (-0.67 * 2a)*(3.18) + c
Simplifying, we have:
0 = 10.1368a - 4.2732a + c
Now we have two equations:
0 = 5.8636a + c
0 = 10.1368a - 4.2732a + c
Solving this system of equations, we find:
a ≈ -5.7244
c ≈ 33.59
Using these values, we can substitute them back into the equation b = -0.67 * 2a:
b ≈ 7.6276
Therefore, the quadratic function f(t) that gives the height of the rock above the road is:
f(t) ≈ -5.7244t^2 + 7.6276t + 33.59.
To define a quadratic function that describes the height of the rock above the road in terms of time, we need to consider the equation of motion for an object moving vertically under the influence of gravity.
The equation of motion can be written as follows:
h(t) = -gt^2 + vt + h0
Where:
h(t) is the height of the object at time t
g is the acceleration due to gravity (approximately -32.2 ft/s^2)
v is the initial vertical velocity of the object
h0 is the initial height of the object
In this case, the rock is thrown upward, so its initial velocity is positive due to the upward direction. Also, the initial height is given as 88 feet above the road.
Using the given information, we can determine the values of g, v, and h0 to define the quadratic function.
1. Finding g:
Since the acceleration due to gravity is negative, we can set g = -32.2 ft/s^2.
2. Finding v:
To find the initial vertical velocity v, we need the maximum height of the rock above the road and the time it takes to reach that height.
Given that the rock reaches its maximum height 0.67 seconds after it is thrown, we can calculate the vertical velocity at that time using the equation:
v = gt - gt_max
where t_max = 0.67 seconds
v = -32.2 ft/s^2 * 0.67 seconds = -21.574 ft/s (approximately)
3. Finding h0:
The initial height of the rock above the road is given as 88 feet. Therefore, h0 = 88 feet.
Now, we can substitute these values into the equation of motion to define the quadratic function:
f(t) = -gt^2 + vt + h0
Plugging in the values we obtained:
f(t) = -32.2t^2 - 21.574t + 88
Therefore, the quadratic function f(t) = -32.2t^2 - 21.574t + 88 gives the height of the rock above the road (in feet) as a function of time t.