Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Let's solve the problem step-by-step.
Part (a): Writing the Model for the Player's Height Above Ground Level
To write the model expressing the player's height [tex]\( s(t) \)[/tex] above ground level [tex]\( t \)[/tex] seconds after leaving the ground, we can use the equation:
[tex]\[ s(t) = -\frac{1}{2} g t^2 + v_0 t + s_0 \][/tex]
where:
- [tex]\( g = 32 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)
- [tex]\( v_0 = 16 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( s_0 = 0 \, \text{ft} \)[/tex] (initial height, since he jumped from ground level)
Substituting these values into the equation, we get:
[tex]\[ s(t) = -\frac{1}{2} (32) t^2 + 16 t + 0 \][/tex]
Simplifying the equation:
[tex]\[ s(t) = -16 t^2 + 16 t \][/tex]
So, the model to express the player's height above ground level [tex]\( t \)[/tex] seconds after leaving the ground is:
[tex]\[ s(t) = -16 t^2 + 16 t \][/tex]
Part (b): Determining the Time to Reach Maximum Height
The maximum height reached by the player is [tex]\( 4 \, \text{ft} \)[/tex] (since [tex]\( 48 \, \text{in} = 4 \, \text{ft} \)[/tex]). We need to determine the time [tex]\( t \)[/tex] it takes for the player to reach this height.
Using the model from part (a):
[tex]\[ s(t) = -16 t^2 + 16 t \][/tex]
we set [tex]\( s(t) \)[/tex] to the maximum height:
[tex]\[ 4 = -16 t^2 + 16 t \][/tex]
To solve for [tex]\( t \)[/tex], we rearrange the equation:
[tex]\[ -16 t^2 + 16 t - 4 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = -4 \)[/tex].
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values:
[tex]\[ t = \frac{-(16) \pm \sqrt{(16)^2 - 4(-16)(-4)}}{2(-16)} \][/tex]
Simplifying further, we get:
[tex]\[ t = \frac{-16 \pm \sqrt{256 - 256}}{-32} \][/tex]
[tex]\[ t = \frac{-16 \pm \sqrt{0}}{-32} \][/tex]
[tex]\[ t = \frac{-16 \pm 0}{-32} \][/tex]
[tex]\[ t = \frac{-16}{-32} \][/tex]
Finally:
[tex]\[ t = 0.5 \][/tex]
So, the time [tex]\( t \)[/tex] it takes for the player to reach his maximum height of 4 feet is [tex]\( 0.5 \)[/tex] seconds.
Part (a): Writing the Model for the Player's Height Above Ground Level
To write the model expressing the player's height [tex]\( s(t) \)[/tex] above ground level [tex]\( t \)[/tex] seconds after leaving the ground, we can use the equation:
[tex]\[ s(t) = -\frac{1}{2} g t^2 + v_0 t + s_0 \][/tex]
where:
- [tex]\( g = 32 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)
- [tex]\( v_0 = 16 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( s_0 = 0 \, \text{ft} \)[/tex] (initial height, since he jumped from ground level)
Substituting these values into the equation, we get:
[tex]\[ s(t) = -\frac{1}{2} (32) t^2 + 16 t + 0 \][/tex]
Simplifying the equation:
[tex]\[ s(t) = -16 t^2 + 16 t \][/tex]
So, the model to express the player's height above ground level [tex]\( t \)[/tex] seconds after leaving the ground is:
[tex]\[ s(t) = -16 t^2 + 16 t \][/tex]
Part (b): Determining the Time to Reach Maximum Height
The maximum height reached by the player is [tex]\( 4 \, \text{ft} \)[/tex] (since [tex]\( 48 \, \text{in} = 4 \, \text{ft} \)[/tex]). We need to determine the time [tex]\( t \)[/tex] it takes for the player to reach this height.
Using the model from part (a):
[tex]\[ s(t) = -16 t^2 + 16 t \][/tex]
we set [tex]\( s(t) \)[/tex] to the maximum height:
[tex]\[ 4 = -16 t^2 + 16 t \][/tex]
To solve for [tex]\( t \)[/tex], we rearrange the equation:
[tex]\[ -16 t^2 + 16 t - 4 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = -4 \)[/tex].
To solve this quadratic equation, we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values:
[tex]\[ t = \frac{-(16) \pm \sqrt{(16)^2 - 4(-16)(-4)}}{2(-16)} \][/tex]
Simplifying further, we get:
[tex]\[ t = \frac{-16 \pm \sqrt{256 - 256}}{-32} \][/tex]
[tex]\[ t = \frac{-16 \pm \sqrt{0}}{-32} \][/tex]
[tex]\[ t = \frac{-16 \pm 0}{-32} \][/tex]
[tex]\[ t = \frac{-16}{-32} \][/tex]
Finally:
[tex]\[ t = 0.5 \][/tex]
So, the time [tex]\( t \)[/tex] it takes for the player to reach his maximum height of 4 feet is [tex]\( 0.5 \)[/tex] seconds.
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.