Answer:
The average velocity for the time period beginning when t=1 and lasting 0.1 seconds = 16.40 ft/s.
Explanation:
Given that the height of the ball at time t is
![\rm h(t) = (50 t-16t^2)\ ft.](https://tex.z-dn.net/?f=%5Crm%20h%28t%29%20%3D%20%2850%20t-16t%5E2%29%5C%20ft.)
The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.
If
are the heights of the ball at time
and
, then the total displacement covered by the ball from time
to
is
.
Thus, the average velocity of the ball for the time interval
is given by
![\rm v_{av}=\dfrac{h_2-h_1}{t_2-t_1}.](https://tex.z-dn.net/?f=%5Crm%20v_%7Bav%7D%3D%5Cdfrac%7Bh_2-h_1%7D%7Bt_2-t_1%7D.)
For the time interval, beginning when t = 1 second and lasting 0.1 seconds,
![\rm t_1=1\ sec.\\t_2 = 1\sec + 0.1\ sec = 1.1\ sec.\\\\h_1=50\times 1-16\times 1^2=34\ ft.\\h_2=50\times 1.1-16\times 1.1^2=35.64\\\\Therefore,\\\\v_{av} = \dfrac{35.64-34}{1.1-1}=16.40\ ft/s.](https://tex.z-dn.net/?f=%5Crm%20t_1%3D1%5C%20sec.%5C%5Ct_2%20%3D%201%5Csec%20%2B%200.1%5C%20sec%20%3D%201.1%5C%20sec.%5C%5C%5C%5Ch_1%3D50%5Ctimes%201-16%5Ctimes%201%5E2%3D34%5C%20ft.%5C%5Ch_2%3D50%5Ctimes%201.1-16%5Ctimes%201.1%5E2%3D35.64%5C%5C%5C%5CTherefore%2C%5C%5C%5C%5Cv_%7Bav%7D%20%3D%20%5Cdfrac%7B35.64-34%7D%7B1.1-1%7D%3D16.40%5C%20ft%2Fs.)