Answer:
The rate of change of height with respect to time is -10.64 feet/sec
Explanation:
Given that,
There are three lines, so you can calculate three different values of the function at one time.
The function f(t) represents the height in feet of a ball thrown into the air, t seconds after it has been thrown.
Given table is,
Time t = 0, 1, 1,02
Function is,![F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015](https://tex.z-dn.net/?f=F%28t%29%3D-3.053113177191196%5Ctimes10%5E%7B-18%7D%2C6.000000000000134%2C%206.41760000000015)
We need to calculate the initial height of ball
Using equation of motion
![h=h_{0}+ut-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dh_%7B0%7D%2But-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Where, h₀ = initial height
u = initial velocity
t = time
g = acceleration due to gravity
At t = 0,
Put the value into the formula
![-3.053113177191196\times10^{-18}=h_{0}+0-0](https://tex.z-dn.net/?f=-3.053113177191196%5Ctimes10%5E%7B-18%7D%3Dh_%7B0%7D%2B0-0)
![h_{0}=-3.053113177191196\times10^{-18}](https://tex.z-dn.net/?f=h_%7B0%7D%3D-3.053113177191196%5Ctimes10%5E%7B-18%7D)
We need to calculate the height of ball at t = 1
Using equation of motion
![h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=h_%7B1%7D%3Dh_%7B0%7D%2Bu_%7B0%7Dt-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
Put the value in the equation
![6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32](https://tex.z-dn.net/?f=6.000000000000134%3D-3.053113177191196%5Ctimes10%5E%7B-18%7D%2Bu-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes32)
![6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}](https://tex.z-dn.net/?f=6.000000000000134%2B3.053113177191196%5Ctimes10%5E%7B-18%7D%2B16%3Du_%7B0%7D)
![u_{0}=22\ feet/s](https://tex.z-dn.net/?f=u_%7B0%7D%3D22%5C%20feet%2Fs)
Velocity is the rate of change of height with respect to time
So, velocity at 1.02 sec is given
We need to calculate the height
Using equation of motion
![h=ut-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dut-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
On differentiating w.r.to t
![h'(t)=u-\dfrac{1}{2}g(2t)](https://tex.z-dn.net/?f=h%27%28t%29%3Du-%5Cdfrac%7B1%7D%7B2%7Dg%282t%29)
Put the value into the formula
![h'(t)=22-\times32\times(1.02)](https://tex.z-dn.net/?f=h%27%28t%29%3D22-%5Ctimes32%5Ctimes%281.02%29)
![h'(t)=-10.64\ feet/sec](https://tex.z-dn.net/?f=h%27%28t%29%3D-10.64%5C%20feet%2Fsec)
Hence, The rate of change of height with respect to time is -10.64 feet/sec.