A diver is on the diving platform at Wonder Mountain in Canada’s Wonderland. She jumps up and dives into the water at the base of the mountain. The equation is d = -3t^2 + 6t + 45
At what time is she 36 m above the water?
When is she less than 21 m from the water?
-3t^2 + 6t + 45 = 36
-3t^2 + 6t + 9 = 0
t^2 - 2t - 3 = 0
(t-3)(t+1) = 0
t=3 or t=-1 , but we know t >= 0
so at 3 seconds she is 36 m above the ground
-3t^2 + 6t + 45 < 21
t^2 - 2t -8 > 0
(t-4)(t+2) > 0
t > 4 or t < -2
let's see when she hits the water.
-3t^2 + 6t + 45 = 0
t^2 - 2t + 15=0
(t-5)(t+3) = 0
t = 5
So she will be between 21 m and the ground
from second 4 and 5 seconds
To find the time at which the diver is 36 m above the water, we need to solve the equation d = 36.
So, substituting d = 36 into the given equation, we get:
36 = -3t^2 + 6t + 45
Rearranging the equation, we have:
-3t^2 + 6t + 45 - 36 = 0
Simplifying further:
-3t^2 + 6t + 9 = 0
Dividing the equation by -3, we get:
t^2 - 2t - 3 = 0
This is a quadratic equation, which can be factored as:
(t - 3)(t + 1) = 0
Setting each factor equal to zero, we have:
t - 3 = 0 or t + 1 = 0
Solving the equations, we find:
t = 3 or t = -1
Since time cannot be negative in this context, we discard t = -1. Therefore, the diver is 36 m above the water at time t = 3.
To find the time at which the diver is less than 21 m from the water, we need to solve the equation d < 21.
Substituting d = 21 into the given equation, we get:
21 = -3t^2 + 6t + 45
Rearranging the equation, we have:
-3t^2 + 6t + 45 - 21 = 0
Simplifying further:
-3t^2 + 6t + 24 = 0
Dividing the equation by -3, we get:
t^2 - 2t - 8 = 0
This is a quadratic equation, which can be factored as:
(t - 4)(t + 2) = 0
Setting each factor equal to zero, we have:
t - 4 = 0 or t + 2 = 0
Solving the equations, we find:
t = 4 or t = -2
Since time cannot be negative in this context, we discard t = -2. Therefore, the diver is less than 21 m from the water at time t = 4.
To find the time at which the diver is 36 m above the water, we need to solve the given equation for 't' when d (distance above the water) is 36 m.
The equation given is d = -3t^2 + 6t + 45.
Step 1: Replace 'd' with 36 in the equation.
36 = -3t^2 + 6t + 45
Step 2: Rearrange the equation in standard quadratic form (set it equal to zero).
0 = -3t^2 + 6t + 45 - 36
0 = -3t^2 + 6t + 9
Step 3: Factor out any common factors, if possible. In this case, we can divide each term by 3 to simplify the equation.
0 = -t^2 + 2t + 3
Step 4: Solve the equation by factoring, completing the square, or using the quadratic formula. In this case, factoring is easiest.
0 = -(t - 1)(t - 3)
Setting each factor equal to zero, we get:
t - 1 = 0 --> t = 1
t - 3 = 0 --> t = 3
Therefore, the diver is 36 m above the water at 1 second and at 3 seconds.
To find the time when the diver is less than 21 m from the water, we need to solve the given equation for 't' when d (distance above the water) is less than 21 m.
Step 1: Replace 'd' with 21 in the equation.
21 = -3t^2 + 6t + 45
Step 2: Rearrange the equation in standard quadratic form (set it equal to zero).
0 = -3t^2 + 6t + 45 - 21
0 = -3t^2 + 6t + 24
Step 3: Divide each term by 3 to simplify the equation.
0 = -t^2 + 2t + 8
Step 4: Solve the equation by factoring, completing the square, or using the quadratic formula. In this case, factoring is simplest.
0 = -(t - 4)(t + 2)
Setting each factor equal to zero, we get:
t - 4 = 0 --> t = 4
t + 2 = 0 --> t = -2
Therefore, the diver is less than 21 m from the water at 4 seconds and at -2 seconds. Note that negative time is not physically meaningful in this context, so we consider only the positive solution.
In summary:
1. The diver is 36 m above the water at 1 second and at 3 seconds.
2. The diver is less than 21 m from the water at 4 seconds.