If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.

Question

(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

Answer 1

see Nick @ 7:10

Answer 2

y(t) = 48t − 16t^2 (note notation)

i)
y(2.5) =48(2.5)-16(6.25) = 20
y(2) = 48(2)-16(4) = 32
20-32 = -12 ft in .5 s
-12/.5 = -24

ii)
y(2.1) = 48*2.1 -16*2.1^2 = 30.24
30.24-32 = -1.76
-1.76/.1 =-17.6

etc getting closer and closer to
dy/dt = v at t = 2 for part b
cheating for that
dy/dt = 48 - 32 t
at t = 2
dy/dt = v = 48 - 64 = -16 :)

Answer 3

@Scott, sketchy lol, Its been since 2018 and I wonder if he's still search for Nick at 7:10

Answer 4

(40(2.5) − 16(2.5)2 − 16)/.5