If a ball is thrown in the air with a velocity 52 ft/s, its height in feet t seconds later is given by y = 52t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting (i) 0.5 second. (ii) 0.1 second. (iii) 0.05 second. (iv) 0.01 second.

More generally, you have after $h$ seconds

$\dfrac{y(2+h)-y(2)}{(2+h)-2}=\dfrac{52(2+h)-16(2+h)^2-104-64}h=\dfrac{-12h-16h^2}h=-12-16h$

After $h=0.5$ second, the average velocity is $-12-16\times0.5=-20$.

I'm sure you can do the rest on your own.

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erinna erinna · Answer 2 · 2019-10-03T04:48:09-04:00

Answer:

(i) $m=-20$

(ii) $m=-13.6$

(iii) $m=-12.8$

(iv) $m=-12.16$

Step-by-step explanation:

The given function is

$y=52t-16t^2$

where, h is the height of ball after t seconds.

It can be written as

$f(t)=52t-16t^2$

At x=2,

$f(2)=52(2)-16(2)^2=40$

The average of a function f(x) on [a,b] is

$m=\frac{f(b)-f(a)}{b-a}$

The time period beginning when t = 2 and lasting with h.

$m=\frac{f(2+h)-f(2)}{2+h-2}$

$m=\frac{f(2+h)-40}{h}$

(i)

Here, h = 0.5 second

$m=\frac{f(2+0.5)-40}{0.5}$

$m=\frac{f(2.5)-40}{0.5}$

$m=\frac{30-40}{0.5}$

$m=-20$

(ii)

Here, h = 0.1 second

$m=\frac{f(2+0.1)-40}{0.1}$

$m=\frac{f(2.1)-40}{0.1}$

$m=\frac{38.64-40}{0.1}$

$m=-13.6$

(iii)

Here, h = 0.05 second

$m=\frac{f(2+0.05)-40}{0.05}$

$m=\frac{f(2.05)-40}{0.05}$

$m=\frac{39.36-40}{0.05}$

$m=-12.8$

(iv)

Here, h = 0.01 second

$m=\frac{f(2+0.1)-40}{0.01}$

$m=\frac{f(2.01)-40}{0.01}$

$m=\frac{39.8784-40}{0.01}$

$m=-12.16$