If a ball is thrown in the air with a velocity 50 ft/s, its height in feet t seconds later is given by y = 50t − 16t^2.

Question

(a) Find the average velocity for the time period beginning when t = 2 and lasting

0.5 second
0.1 second
0.05 second
0.01 second

i plugged in 50(0.5)-16(0.5)^2 i did not get the answer -22...

Answer 1

I will do the first one:
when t = 2
y = 50(2) - 16(4) = 36
when t = 2.5
y = 50(2.5) - 16(2.5)^2 = 25

velocity = (25 - 36)/(2.5 - 2)
= -22

repeat for the others, remember you start with t = 2 and add on the given value
so for the next one, t =2 and t = 2.1

Answer 2

First you find the height after 2 seconds, we'll denote that as y2. So using the formula we have:

y2=50.2-16.2^2
-> 100-16.4=100-64=36f
After 0.5 seconds, the time passed after the throw will be 2.5 seconds so for this height we have:

y2.5=50*0.5-16*(0.5)^2
-> 25-4=21
So after 2.5 second the height of the ball will be 21f

We know that the average velocity can be calculated using the following formula:

v=Δs/Δt=|y1−y0|/|t1−t0|
v=Δs/Δt=|y1−y0|/|t1−t0|
So after the substitution we have:

v=|21-36|/|2.5−2|= |-5|/|0.5|=5/0.5
=5*10/5=10

So the average speed in that period will be 10 ftsfts
Now you can do the others by yourself.

Answer 3

sorry my 2.5 is wrong its 25 i incorrectly put 0.5 for 2.5 but rest can be corrected by that and u get the correct ans not like me :) sorry