"A rock is dropped from the edge of a 180 m cliff. the function f(x) = -5t^2 - 5t + 180 gives the height of the rock, h(t), in meters t seconds after it was released. How long does it take for the rock to reach a ledge 80 m above the base of the cliff?"
I have no idea how to solve this, steps and explanations would be amazing.
well, since f(t) is the height of the rock at time t, just set it to 80 and solve for t:
-5t^2 - 5t + 180 = 80
5t^2 + 5t - 100 = 0
t^2 + t - 20 = 0
(t+5)(t-4) = 20
t=4
check:
f(4) = -5(16)-5(4)+180 = 80
W we tp
To solve this problem, we need to find the time it takes for the rock to reach a ledge 80 m above the base of the cliff.
Let's set up the equation:
h(t) = -5t^2 - 5t + 180
We want to find the time when the height is equal to 80 m:
h(t) = 80
Now we can substitute the height into the equation:
-5t^2 - 5t + 180 = 80
Rearrange the equation to make it equal to zero:
-5t^2 - 5t + 100 = 0
Next, we can solve this quadratic equation using either factoring, completing the square, or the quadratic formula. In this case, we will use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -5, b = -5, and c = 100. Plug in these values into the quadratic formula:
t = (-(-5) ± √((-5)^2 - 4(-5)(100))) / (2(-5))
Simplify the equation:
t = (5 ± √(25 + 2000)) / (-10)
t = (5 ± √2025) / (-10)
Since we can't have a negative time in this context, we take the positive root:
t = (5 + √2025) / (-10)
Now, we can calculate the square root:
t = (5 + 45) / (-10)
t = 50 / (-10)
Simplify:
t = -5
So, it takes the rock 5 seconds to reach the ledge 80 m above the base of the cliff.
To find the time it takes for the rock to reach a ledge 80 m above the base of the cliff, we need to set up an equation based on the given information.
The function f(x) = -5t^2 - 5t + 180 gives the height of the rock, h(t), in meters t seconds after it was released.
Let's call the time it takes for the rock to reach the ledge as "t_ledge". At this time, the height of the rock will be 80 meters above the base of the cliff.
We can set up an equation by equating the height of the rock at time t_ledge with 80 m and solve for t_ledge.
So, we have: -5t_ledge^2 - 5t_ledge + 180 = 80
To solve this equation, let's rearrange it to the standard form of a quadratic equation: -5t_ledge^2 - 5t_ledge + 100 = 0
Now, we can solve this equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -5, b = -5, and c = 100.
Applying the quadratic formula, we get:
t = (-(-5) ± √((-5)^2 - 4*(-5)*100)) / (2*(-5))
t = (5 ± √(25 + 2000)) / -10
t = (5 ± √2025) / -10
t = (5 ± 45) / -10
So, we have two possible solutions for t_ledge:
1. t_ledge = (5 + 45) / -10 = 50 / -10 = -5 (Not possible since time cannot be negative)
2. t_ledge = (5 - 45) / -10 = -40 / -10 = 4
The rock takes 4 seconds to reach the ledge 80 m above the base of the cliff.