Answer with Step-by-step explanation:
We are given that
Height (in feet) of ball
[tex]y=36t-16t^2[/tex]
a.We have to find the average velocity for the time period t=2 and lasting
(i) 0.5 s
Therefore, t=2+0.5=2.5 s
Average of function on interval [a,b]=f'(c)=[tex]\frac{f(b)-f(a)}{b-a}[/tex]
[tex]y(2)=36(2)-16(2)^2=72-64=8[/tex]
[tex]y(2.5)=36(2.5)-16(2.5)^2=-10[/tex]
Using the formula
Average velocity on interval [2,2.5]=[tex]v(t)=\frac{-10-8}{2.5-2}=\frac{-18}{0.5}=-36ft/s[/tex]
(ii) 0.1 s
t=2+0.1=2.1 s
[tex]y(2.1)=36(2.1)-16(2.1)^2=5.04[/tex]
Average velocity on interval [2,2.1]=[tex]v(t)=\frac{5.04-8}{2.1-2}=-29.6ft/s[/tex]
(iii) 0.05 s
t=2+0.05=2.05 s
[tex]y(2.05)=36(2.05)-16(2.05)^2=6.56[/tex]
Average velocity on interval [2,2.05]=[tex]v(t)=\frac{6.56-8}{2.05-2}=-28.8ft/s[/tex]
(iv) 0.01 s
t=2+0.01=2.01 s
[tex]y(2.01)=36(2.01)-16(2.01)^2=7.7184[/tex]
Average velocity on interval [2,2.01]=[tex]v(t)=\frac{7.7184-8}{2.01-2}[/tex]
Average velocity on interval [2,2.01]=[tex]-28.16 ft/s[/tex]
(b) t=2 s
We have to find the instantaneous velocity at t=2 s
Instantaneous velocity=[tex]v(t)=\frac{ds}{dt}[/tex]
Using the formula
[tex]v(t)=\frac{d(36t-16t^2)}{dt}=36-32t[/tex]
Substitute t=2
Instantaneous velocity at t=2 s
v(2)=36-32(2)=-28 ft/s
