A farmer climbs to the top of a silo, which is 80 feet high, and throws a rock straight up at a speed of 64 ft/sec. The height of the rock in feet above the ground after t seconds is given by the function h(t)=-16t^2+64t+80.
When will the rock hit the ground?
When will the rock be at its highest point? What is the highest point?
To find when the rock will hit the ground, we need to find the value of t when h(t) = 0. So we need to solve the equation -16t^2 + 64t + 80 = 0.
To find when the rock will be at its highest point, we need to find the vertex of the quadratic function. The x-coordinate of the vertex is found using the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic function. In this case, a = -16 and b = 64.
x = -64/2*-16
x = -64/-32
x = 2
The highest point occurs after 2 seconds.
To find the highest point, we substitute t = 2 into the function h(t):
h(2) = -16(2)^2 + 64(2) + 80
h(2) = -64 + 128 + 80
h(2) = 144
The highest point is at a height of 144 feet.
To find when the rock will hit the ground, we need to set the height function h(t) equal to 0 and solve for t:
-16t^2 + 64t + 80 = 0
We can simplify this equation by dividing by -16:
t^2 - 4t - 5 = 0
Next, we can factor the quadratic equation:
(t - 5)(t + 1) = 0
Setting each factor equal to zero gives us two possible values for t:
t - 5 = 0 --> t = 5
t + 1 = 0 --> t = -1
Since time cannot be negative in this context, we can disregard t = -1. Therefore, the rock will hit the ground after 5 seconds.
To find when the rock will be at its highest point, we can note that the height function h(t) is a quadratic equation in the form of h(t) = -16t^2 + 64t + 80. The graph of a quadratic function is a parabola, and the highest point of a parabola occurs at its vertex.
The t-coordinate of the vertex can be found using the formula t = -b/(2a), where a, b, and c are the coefficients of the quadratic equation. In our case, a = -16 and b = 64.
t = -64 / (2 * -16)
t = -64 / -32
t = 2
Therefore, the rock will be at its highest point 2 seconds after it is thrown.
To find the highest point, we can substitute t = 2 into the height function h(t):
h(2) = -16(2)^2 + 64(2) + 80
h(2) = -16(4) + 128 + 80
h(2) = -64 + 128 + 80
h(2) = 144
Therefore, the highest point of the rock is 144 feet above the ground.
To find when the rock will hit the ground, we need to determine the value of t when h(t) equals 0. In this case, h(t) represents the height of the rock above the ground.
So, we set h(t) equal to 0 and solve the equation:
-16t^2 + 64t + 80 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -16, b = 64, and c = 80.
Applying these values to the formula, we have:
t = (-64 ± √(64^2 - 4(-16)(80))) / (2(-16))
Simplifying further:
t = (-64 ± √(4096 + 5120)) / (-32)
t = (-64 ± √9216) / (-32)
t = (-64 ± 96) / (-32)
We have two possible solutions:
1. t = (-64 + 96) / (-32) = 32 / (-32) = -1
2. t = (-64 - 96) / (-32) = -160 / (-32) = 5
Since time cannot be negative in this context, we discard the first solution (-1). Therefore, the rock will hit the ground after 5 seconds.
To find when the rock will be at its highest point, we need to determine the maximum point of the quadratic function h(t). The highest point of a quadratic function occurs at the vertex.
The formula for the x-coordinate of the vertex of a quadratic function is given by:
x = -b / (2a)
In this case, a = -16 and b = 64. Plugging in these values, we get:
x = -64 / (2(-16)) = -64 / (-32) = 2
So, the rock will be at its highest point after 2 seconds.
Now, to find the highest point, we substitute the value of t (2 seconds) back into the function h(t):
h(2) = -16(2)^2 + 64(2) + 80
Simplifying further:
h(2) = -16(4) + 128 + 80
h(2) = -64 + 128 + 80
h(2) = 144
Therefore, the highest point of the rock's trajectory is 144 feet above the ground.