Question

If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it height in feet after t second is given by y=45t−16t2. Find the average velocity for the time period begining when t=2 and lasting. 0.1 seconds?

Answer 1

The average velocity of the ball between t = 2 and t = 2.1 seconds is calculated using the change in position over the change in time and is found to be -20.6 ft/s.

Step-by-step explanation:

To calculate the average velocity of the ball over the time interval from t = 2 seconds to t = 2.1 seconds, we use the given position function y = 45t - 16t2. The average velocity is the change in position divided by the change in time. We can find the position of the ball at t = 2 seconds and at t = 2.1 seconds and then compute the difference.

At t = 2 seconds: y = 45(2) - 16(2)2 = 90 - 64 = 26 feet

At t = 2.1 seconds: y = 45(2.1) - 16(2.1)2 = 94.5 - 70.56 = 23.94 feet

The change in position Δy is 23.94 feet - 26 feet = -2.06 feet (the ball is going down as the difference is negative). The change in time Δt is 2.1 seconds - 2 seconds = 0.1 seconds. So, the average velocity vavg over the interval from t = 2 to t = 2.1 seconds is:

vavg = Δy / Δt = -2.06 feet / 0.1 seconds = -20.6 ft/s

Answer 2

y = 45 t - 16 t²
f ( 2 s ) = 45 · 2 - 16 · 4 = 90 - 64 = 26 ft
f ( 2.1 s ) = 45 · 2.1 - 16 · 4.41 = 94.5 - 70.56 = 23.94 ft
Average velocity:
v = ( 23.94 - 26 ) / ( 2.1 - 2 ) = - 2.06 ft / 0.1 s = - 20.6 ft/s