Answer :
Certainly! Let's address the questions step by step:
### (a) Domain and Range of h(t)
The function given is [tex]\( h(t) = -0.5t^2 + v_0t + h_0 \)[/tex] with [tex]\( v_0 = 0 \)[/tex] m/s and [tex]\( h_0 = 210 \)[/tex] meters.
Domain: The domain of [tex]\( h(t) \)[/tex] in this physical scenario represents the time for which the height calculation is valid. Since the time starts when the jump begins and extends forward, the domain is [tex]\( t \in [0, \infty) \)[/tex]. This signifies that we start measuring time from the moment the jumper leaps off the bridge and look into the future.
Range: The range of [tex]\( h(t) \)[/tex] is the set of possible heights the jumper can have. The height starts from the initial height [tex]\( h_0 = 210 \)[/tex] meters and can go down to 0 meters (the river surface). Therefore, the range is [tex]\( [0, 210] \)[/tex] meters. This signifies that the lowest height the jumper can reach is 0 meters and the maximum height is 210 meters.
### (b) Vertex of the Height Function
The vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a < 0 \)[/tex], represents its maximum value. Here, we seek the vertex of [tex]\( h(t) = -0.5t^2 + 210 \)[/tex].
The time component [tex]\( t \)[/tex] where the vertex (maximum height) occurs is given by the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
For the given scenario:
- [tex]\( a = -0.5 \)[/tex]
- [tex]\( b = v_0 = 0 \)[/tex]
So,
[tex]\[ t = -\frac{0}{2 \cdot -0.5} = 0 \][/tex]
The vertex in this context represents the initial conditions, so the height at [tex]\( t = 0 \)[/tex] is:
[tex]\[ h(0) = 210 \text{ meters} \][/tex]
Thus, the vertex is [tex]\( (0, 210) \)[/tex]. It signifies that the maximum height of 210 meters is at the initial time [tex]\( t = 0 \)[/tex].
### (c) Maximum Height and Time
Since we have already determined the vertex, we know the jumper starts at the maximum height. Therefore:
- Maximum height is [tex]\( 210 \)[/tex] meters.
- Time of the maximum height is [tex]\( t = 0 \)[/tex] seconds.
Given there is no initial upward velocity, the max height is indeed the starting height.
### (d) Time to Reach Height of 11m
To determine the time when [tex]\( h(t) = 11 \)[/tex] meters, we need to solve the quadratic equation:
[tex]\[ -0.5t^2 + 210 = 11 \][/tex]
Rearranging, we get:
[tex]\[ -0.5t^2 + 199 = 0 \][/tex]
This is a standard quadratic equation [tex]\( at^2 + bt + c = 0 \)[/tex] form:
[tex]\[ a = -0.5, \quad b = 0, \quad c = 199 \][/tex]
The discriminant [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[ \Delta = 0^2 - 4(-0.5)(199) = 398 \][/tex]
Since the discriminant is positive, there are two potential times:
[tex]\[ t = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{0 \pm \sqrt{398}}{-1} \][/tex]
This simplifies to:
[tex]\[ t_1 = \sqrt{398} \approx 19.95 \text{ seconds} \][/tex]
[tex]\[ t_2 = -\sqrt{398} \quad (\text{not considered as } t \geq 0) \][/tex]
However, for practical purposes, the jump doesn't actually reach 11 meters as the detailed quadratic analysis yields no real positive solution fitting within the physical constraints seen in the results from the tool.
### (e) Height After 20 Seconds
To find the height at [tex]\( t = 20 \)[/tex] seconds:
[tex]\[ h(20) = -0.5(20)^2 + 210 \][/tex]
[tex]\[ h(20) = -0.5(400) + 210 \][/tex]
[tex]\[ h(20) = -200 + 210 = 10 \text{ meters} \][/tex]
Thus, the height after 20 seconds is 2170 meters, signifying an incorrect height suggesting the assumed scenario does not maintain physical realism.
### (f) Time to Touch the River
This is when [tex]\( h(t) = 0 \)[/tex]. We solve:
[tex]\[ -0.5t^2 + 210 = 0 \][/tex]
[tex]\[ -0.5t^2 = -210 \][/tex]
[tex]\[ t^2 = 420 \][/tex]
[tex]\[ t = \sqrt{420} \approx 20.49 \text{ seconds} \][/tex]
Thus, the time to touch the river is [tex]\( 20.49 \)[/tex] seconds.
### (a) Domain and Range of h(t)
The function given is [tex]\( h(t) = -0.5t^2 + v_0t + h_0 \)[/tex] with [tex]\( v_0 = 0 \)[/tex] m/s and [tex]\( h_0 = 210 \)[/tex] meters.
Domain: The domain of [tex]\( h(t) \)[/tex] in this physical scenario represents the time for which the height calculation is valid. Since the time starts when the jump begins and extends forward, the domain is [tex]\( t \in [0, \infty) \)[/tex]. This signifies that we start measuring time from the moment the jumper leaps off the bridge and look into the future.
Range: The range of [tex]\( h(t) \)[/tex] is the set of possible heights the jumper can have. The height starts from the initial height [tex]\( h_0 = 210 \)[/tex] meters and can go down to 0 meters (the river surface). Therefore, the range is [tex]\( [0, 210] \)[/tex] meters. This signifies that the lowest height the jumper can reach is 0 meters and the maximum height is 210 meters.
### (b) Vertex of the Height Function
The vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a < 0 \)[/tex], represents its maximum value. Here, we seek the vertex of [tex]\( h(t) = -0.5t^2 + 210 \)[/tex].
The time component [tex]\( t \)[/tex] where the vertex (maximum height) occurs is given by the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
For the given scenario:
- [tex]\( a = -0.5 \)[/tex]
- [tex]\( b = v_0 = 0 \)[/tex]
So,
[tex]\[ t = -\frac{0}{2 \cdot -0.5} = 0 \][/tex]
The vertex in this context represents the initial conditions, so the height at [tex]\( t = 0 \)[/tex] is:
[tex]\[ h(0) = 210 \text{ meters} \][/tex]
Thus, the vertex is [tex]\( (0, 210) \)[/tex]. It signifies that the maximum height of 210 meters is at the initial time [tex]\( t = 0 \)[/tex].
### (c) Maximum Height and Time
Since we have already determined the vertex, we know the jumper starts at the maximum height. Therefore:
- Maximum height is [tex]\( 210 \)[/tex] meters.
- Time of the maximum height is [tex]\( t = 0 \)[/tex] seconds.
Given there is no initial upward velocity, the max height is indeed the starting height.
### (d) Time to Reach Height of 11m
To determine the time when [tex]\( h(t) = 11 \)[/tex] meters, we need to solve the quadratic equation:
[tex]\[ -0.5t^2 + 210 = 11 \][/tex]
Rearranging, we get:
[tex]\[ -0.5t^2 + 199 = 0 \][/tex]
This is a standard quadratic equation [tex]\( at^2 + bt + c = 0 \)[/tex] form:
[tex]\[ a = -0.5, \quad b = 0, \quad c = 199 \][/tex]
The discriminant [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[ \Delta = 0^2 - 4(-0.5)(199) = 398 \][/tex]
Since the discriminant is positive, there are two potential times:
[tex]\[ t = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{0 \pm \sqrt{398}}{-1} \][/tex]
This simplifies to:
[tex]\[ t_1 = \sqrt{398} \approx 19.95 \text{ seconds} \][/tex]
[tex]\[ t_2 = -\sqrt{398} \quad (\text{not considered as } t \geq 0) \][/tex]
However, for practical purposes, the jump doesn't actually reach 11 meters as the detailed quadratic analysis yields no real positive solution fitting within the physical constraints seen in the results from the tool.
### (e) Height After 20 Seconds
To find the height at [tex]\( t = 20 \)[/tex] seconds:
[tex]\[ h(20) = -0.5(20)^2 + 210 \][/tex]
[tex]\[ h(20) = -0.5(400) + 210 \][/tex]
[tex]\[ h(20) = -200 + 210 = 10 \text{ meters} \][/tex]
Thus, the height after 20 seconds is 2170 meters, signifying an incorrect height suggesting the assumed scenario does not maintain physical realism.
### (f) Time to Touch the River
This is when [tex]\( h(t) = 0 \)[/tex]. We solve:
[tex]\[ -0.5t^2 + 210 = 0 \][/tex]
[tex]\[ -0.5t^2 = -210 \][/tex]
[tex]\[ t^2 = 420 \][/tex]
[tex]\[ t = \sqrt{420} \approx 20.49 \text{ seconds} \][/tex]
Thus, the time to touch the river is [tex]\( 20.49 \)[/tex] seconds.
