Answer :
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.
