If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet t seconds later is given by y = 40t - 16t^2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting (1) 0.5 second (2) 0.1 second (3) 0.05 second (4) 0.01 second
(b) Estimate the instantaneous velocity when t = 2
When the ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by y = 40t - 16t^2.
The height of the ball after 2 seconds is equal to 40*2 - 16*4 = 16 feet.
(1) Its height after 2.5 seconds is 0. As the ball travels 16 feet in 0.5 seconds, its average velocity for this duration of time is 32 ft/s in the downward direction.
(2) Its height after 2.1 second is 13.44 feet. The ball travels 2.56 feet in 0.1 seconds. It average velocity is 25.6 ft/s in the downward direction.
(3) Its height after 0.05 seconds is 14.76 feet. The ball travels 1.24 feet in 0.05 seconds. Its average velocity is 24.8 feet/second in the downward direction.
(4) Its height after 2.01 second is 15.7584 feet. The distance traveled by the ball in 0.01 second is 0.2416 feet. Its average velocity is 24.16 feet/s in the downward direction.
The instantaneous velocity of the ball when t = 2 is equal to `(d(40t-16t^2))/(dt)` at t = 2. This is 40 - 32*2 = -24 feet/second. The negative sign indicates velocity in the downward direction.