Respuesta :
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,[tex]F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015[/tex]
We need to calculate the initial height of ball
Using equation of motion
[tex]h=h_{0}+ut-\dfrac{1}{2}gt^2[/tex]
Where, hâ‚€ = initial height
u = initial velocity
t = time
g = acceleration due to gravity
At t = 0,
Put the value into the formula
[tex]-3.053113177191196\times10^{-18}=h_{0}+0-0[/tex]
[tex]h_{0}=-3.053113177191196\times10^{-18}[/tex]
We need to calculate the height of ball at t = 1
Using equation of motion
[tex]h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2[/tex]
Put the value in the equation
[tex]6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32[/tex]
[tex]6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}[/tex]
[tex]u_{0}=22\ feet/s[/tex]
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
[tex]h=ut-\dfrac{1}{2}gt^2[/tex]
On differentiating w.r.to t
[tex]h'(t)=u-\dfrac{1}{2}g(2t)[/tex]
Put the value into the formula
[tex]h'(t)=22-\times32\times(1.02)[/tex]
[tex]h'(t)=-10.64\ feet/sec[/tex]
Hence, The rate of change of height with respect to time is -10.64 feet/sec.
