If a ball is thrown straight up into the air with an initial velocity of 45 ft/s

Question

If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it's height in feet after t second is given by y = 45t - 16t^2

a] find the average velocity for the time period begining when t = 2 and lasting.

i) 0.1 seconds

ii) 0.001 seconds

iii) 0.0001 seconds

b) Estimate the instantaneous velocity of the ball when t = 2

Answer 1

velocity = v = y' = 45-32t

v(2) = 45-32(2) =45-64 = -19 ft per second

= 19 ft per second in a downward direction

negative velocity means it's falling back to the ground

v(2.1) = 45-32(2.1) = 45-67.2 = -22.2 ft per second

average velocity from t=2 to t=2.1 = (-19-22.2)/2 = -41.2/2 =-20.5 ft/sec

v(2.001) = 45-32(2.001) = 45-64.032 = -19.032.

average from t=2 to 2.001 is (-19-19.032)/2 = -19.016 ft/sec

v(2.0001) = 45-32(2.0001) =45-64.0032 = -19.0032

average v from t=2 to 2.0001 is (-19-19.0032)/2 = -19.0016 ft/sec

instantaneous v at t=2 = -19 ft/sec