Question

A ball is thrown straight up into the air with an initial velocity of 50 ft/sec. The height h(t) of the ball after t seconds is given by LaTeX: h\left(t\right)=50y-16t^2h ( t ) = 50 y − 16 t 2. What is the average velocity for the time period beginning when t=1 and lasting 0.1 seconds?

Answer 1

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

$\rm h(t) = (50 t-16t^2)\ ft.$

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 $\rm h_1,\ h_2$ are the heights of the ball at time $\rm t_1$ and $\rm t_2$, then the total displacement covered by the ball from time $\rm t_2$ to $\rm t_1$ is $\rm h_2-h_1$.

Thus, the average velocity of the ball for the time interval $\rm t_2-t_1$  is given by

$\rm v_{av}=\dfrac{h_2-h_1}{t_2-t_1}.$

For the time interval, beginning when t = 1 second and lasting 0.1 seconds,

$\rm t_1=1\ sec.\\t_2 = 1\sec + 0.1\ sec = 1.1\ sec.\\\\h_1=50\times 1-16\times 1^2=34\ ft.\\h_2=50\times 1.1-16\times 1.1^2=35.64\\\\Therefore,\\\\v_{av} = \dfrac{35.64-34}{1.1-1}=16.40\ ft/s.$