"if a ball is thrown into the air with a velocity of 44 ft/s, its height in feet t seconds later is given by y = 44t − 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 ft/s (ii) 0.1 seconds ft/s (iii) 0.05 seconds ft/s (iv) 0.01 seconds ft/s (b) estimate the instantaneous velocity when t = 2. ft/s"(c) find the average velocity for the time period beginning whent = 2 and lasting 0.05 second. ft/s(d) find the average velocity for the time period beginning whent = 2 and lasting 0.01 second. ft/s(e) estimate the instantaneous velocity when t = 2.

i) -28 ft/s

ii) -21.6 ft/s

iii) -20.8 ft/s

iv) -20.16 ft/s

b)  -20 ft/s

Step-by-step explanation:

if v represents vertical velocity then if the height y is

y = 44*t − 16*t²

and the instant velocity v is the derivative with respect to the time

v= dy/dt = 44*(1) - 16*(2*t) = 44- 32*t

while the average velocity is va= (y-y₀)/(t-t₀)

where t₀ = 2 and y₀=y(t₀) =  44*2 − 16*2² = 24

then

va= (y-y₀)/(t-t₀) = (44*t − 16*t² -  24)/(t-2)

for

i) t= 0.5s + 2 s= 2.5 s

va = (44* 2.5  − 16* 2.5 ² -  24)/( 2.5 -2) = -28 ft/s

ii) t= 0.1s + 2 s= 2.1 s

va = (44* 2.1  − 16* 2.1 ² -  24)/( 2.1 -2) = -21.6 ft/s

iii) t= 0.05 s + 2 s= 2.05 s

va = (44* 2.05  − 16* 2.05 ² -  24)/( 2.05 -2) = -20.8 ft/s

iv) t= 0.01s + 2 s= 2.01 s

va = (44* 2.01  − 16* 2.01 ² -  24)/( 2.01 -2) = -20.16 ft/s

b) the instantaneous velocity when t=2

v (t=2) =  44- 32*(2) = -20 ft/s

domain is all the x values included in the

0 ≤ x ≤ 2

range is all the y values of the graph

-2 ≤ y ≤ 2

***keep in mind that there are different ways to show domain and range

hope this !

~just a girl in love with shawn mendes

i think the answer is b.

step-by-step explanation:

not 100% sure though.