Question

A bowling ball dropped from a height of $400$feet will be $s\left(t\right)=400-16t^2$feet from the ground after $t$seconds. Use a sequence of average velocities to estimate the instantaneous velocities described below:

When the bowling ball is first dropped, with $h=0.5,$$h=0.25,andh=0.1$

Short answer

Ans: When $h=0.5$instantaneous velocity is: $-8$

When $h=0.25$instantaneous velocity is: $-4$

When $h=0.1$instantaneous velocity is:$-1.6$

Step-by-step solution

Step 1. Given information.

given,

A ball is dropped from a height of $400$feet and its distance from the ground after $t$seconds is, $s\left(t\right)=400-16t^2$

The objective is to estimate the instantaneous velocities for,

$h=0.5,h=0.25,h=0.1$

Step 2. When the ball has just dropped the value of c=0.

To find the instantaneous velocity for $h=0.5$follows the steps:

Now,

$$\begin{gathered} f\left(c\right)=f\left(0\right) \\ =400 \end{gathered}$$

And

$f(c+h)=f(0.5)=396$

therefore the instantaneous velocity is:

$\frac{f(c+h)-f\left(c\right)}{h}=\frac{396-400}{0.5}=-8$

Step 3. To find the instantaneous velocity for h=0.25 follows the steps:

$f\left(c\right)=f\left(0\right)=400$

And,

$f(c+h)=f(0.25)=399$

therefore the instantaneous velocity is:

$\frac{f(c+h)-f\left(c\right)}{h}=\frac{399-400}{0.25}=-4$

$f(c+h)=f(0.25)=399$

Step 4. To find the instantaneous velocity for h=0.1 follows the steps:

$f\left(c\right)=f\left(0\right)=400$

And,

$f(c+h)=f(0.1)=399.84$

therefore the instantaneous velocity is:

$\frac{f(c+h)-f\left(c\right)}{h}=\frac{399.84-400}{0.1}=-1.6$