Question

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
(ii) 0.1 seconds
(iii) 0.05 seconds
(iv) 0.01 seconds

Answer 1

Given:
u = 44 ft/s, the initial upward velocity of the ball

The height in feet after t seconds is
y = 44t - 16t²

Initial time is t₁ = 2 s.
The height at 2 s is
h₁ = 44*2 - 16*(2²) = 24 ft

(i) When the duration is 0.5 s, the final time is
    t₂ = 2 +0.5 = 2.5
    The height at 2.5 s is
     h₂ = 44*2.5 - 16(2.5²) = 10 ft
    The average velocity is
    v = (10 - 24 ft)/(0.5 s) = -28 ft/s
    Answer: - 28 ft/s

(ii) When the duration is 001s, then t₂ = 2.1 s
    h₂ = 44*2.1 - 16(2.1²) = 21.84 ft
    The average velocity is
    v = (21.84 - 24)/0.1 = -21.6 ft/s
   Answer: - 21.6 ft/s

(iii) When the duration is 0.05s, t₂ = 2.05 s
    h₂ = 44*2.05 - 16(2.05²) = 22.96
   Average velocity is
   v = (22.96 - 24)/.05 = - 20.8 ft/s
   Answer:  - 20.8 ft/s

(iv) When the duration is 0.01 s, t₂ = 2.01 s
  h₂ = 44*201 - 16(2.01²) = 23.7984 ft
   v = (23.7984 - 24)/0.01 = - 20.16 ft/s
  Answer: - 20.16 ft/s

As the duration gets smaller, the computed average velocity approaches the true value at 2 s, because it is the derivative of y with respect to t.