# If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet seconds later is given by y = 40t -16t^2.Find the average velocity for the time period beginning when t = 2 and lasting ,0.5 second ,0.1 second, 0.05 second and 0.01 second. What is the instantaneous velocity when t = 2.

A ball is thrown into the air with an initial velocity of 40 ft/s vertically upwards. t seconds after it has been thrown up, the height of the ball is given by the function `f(t) = 40t - 16t^2` .

The average velocity of the ball for a time period between `t_0` and `t_1` is given by:

v = `(f(t_1) - f(t_0))/(t_1 - t_0)`

To measure the average velocity for the time period beginning t = 2, `t_0` = 2.

If it lasts 0.5 s, `t_1` = 2.5:

average velocity = `(40*2.5 - 16*2.5^2 - 40*2 + 16*2^2)/(0.5) = -32`

If it lasts for 0.1 s, `t_1` = 2.1

average velocity = `(40*2.1 - 16*2.1^2 - 40*2 + 16*2^2)/(0.1) = -25.6`

If it lasts for 0.05 s, `t_1` = 2.05

average velocity = `(40*2.05 - 16*2.05^2 - 40*2 + 16*2^2)/(0.05) = -24.8`

If it lasts for 0.01 s, `t_1` = 2.01

average velocity = `(40*2.01 - 16*2.01^2 - 40*2 + 16*2^2)/(0.01) = -24.16`

The instantaneous velocity at t = 2, is given by the value of the derivative of f(t) = 40t - 16t^2 at t = 2.

f'(t) = 40 - 32t

f'(2) = 40 - 64 = -24 ft/s