Question

A ball thrown straight upward from a height of 4 ft with an initial velocity of 50 ft per sec has height h(t) feet after t seconds, where h(t)= - 16t^2 + 50t + 4. What is the ball’s average velocity (that is, its average rate of change in height) during the first second?The average velocity during the first second is ___ feet per second. (Type an integer or a simplified fraction.)

Answer 1

Answer:

34 ft/s

Explanation:

The average rate of change in height during the first second can be calculated as

$\text{ Avg rate = }\frac{h(1)-h(0)}{1-0}$

Because the time goes from 0 to 1 second.

Using the given equation for h(t), we can calculate h(1) and h(0) as follows

$\begin{gathered} h(t)=-16t^2+50t+4 \\ h(1)=-16(1)^2+50(1)+4 \\ h(1)=-16+50+4 \\ h(1)=38 \\ \\ h(0)=-16(0)^2+50(0)+4 \\ h(0)=0+0+4 \\ h(0)=4 \end{gathered}$

Then, the average rate of change in height is

$\text{ Avg rate = }\frac{38-4}{1-0}=\frac{34}{1}=34$

Therefore, the average velocity during the first second is 34 ft/s