Question

If a ball is thrown into the air with a velocity of 36 ft/s, its height in feet t seconds later is given by y = 36t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following. (i) 0.5 seconds 28 Incorrect: Your answer is incorrect. ft/s (ii) 0.1 seconds ft/s (iii) 0.05 seconds ft/s(iv) 0.01 second.(v) t = 2.

Answer 1

Answer with Step-by-step explanation:

We are given that

Height (in feet) of ball

`y=36t-16t^2`

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)=
`(f(b)-f(a))/(b-a)`

`y(2)=36(2)-16(2)^2=72-64=8`

`y(2.5)=36(2.5)-16(2.5)^2=-10`

Using the formula

Average velocity on interval [2,2.5]=
`v(t)=(-10-8)/(2.5-2)=(-18)/(0.5)=-36ft/s`

(ii) 0.1 s

t=2+0.1=2.1 s

`y(2.1)=36(2.1)-16(2.1)^2=5.04`

Average velocity on interval [2,2.1]=
`v(t)=(5.04-8)/(2.1-2)=-29.6ft/s`

(iii) 0.05 s

t=2+0.05=2.05 s

`y(2.05)=36(2.05)-16(2.05)^2=6.56`

Average velocity on interval [2,2.05]=
`v(t)=(6.56-8)/(2.05-2)=-28.8ft/s`

(iv) 0.01 s

t=2+0.01=2.01 s

`y(2.01)=36(2.01)-16(2.01)^2=7.7184`

Average velocity on interval [2,2.01]=
`v(t)=(7.7184-8)/(2.01-2)`

Average velocity on interval [2,2.01]=
`-28.16 ft/s`

(b) t=2 s

We have to find the instantaneous velocity at t=2 s

Instantaneous velocity=
`v(t)=(ds)/(dt)`

Using the formula

`v(t)=(d(36t-16t^2))/(dt)=36-32t`

Substitute t=2

Instantaneous velocity at t=2 s

v(2)=36-32(2)=-28 ft/s