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.
(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...
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.
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
Ah, the wonders of physics equations. Don't worry, I'm here to help you out with your average velocity conundrum.
To find the average velocity, we need to calculate the change in height over the given time period. Let's start with the first time period:
Time period: 0.5 second
To find the average velocity, we need to find the change in height. So, let's find the height at the beginning and at the end of the time period:
At t=2 seconds: y = 50(2) - 16(2)^2 = 20 feet
At t=2.5 seconds: y = 50(2.5) - 16(2.5)^2 = 21.25 feet
Now, let's calculate the change in height:
Change in height = 21.25 feet - 20 feet = 1.25 feet
To find the average velocity, we divide the change in height by the time period:
Average velocity = Change in height / Time period
Average velocity = 1.25 feet / 0.5 seconds = 2.5 ft/s
So, for a time period of 0.5 second, the average velocity is 2.5 ft/s.
Now, it's your turn to tackle the other time periods. You got this!
To find the average velocity for a given time period, you need to calculate the change in height (y) over the change in time (t). Let's go through the steps to find the average velocity for each of the given time periods.
(a) For the time period of 0.5 seconds:
The initial time (t1) is 2 seconds, and the final time (t2) is 2 + 0.5 = 2.5 seconds.
To calculate the change in height (Δy), substitute t1 into the equation and obtain the initial height, y1.
Then substitute t2 into the equation and obtain the final height, y2.
Finally, calculate the average velocity using the formula: average velocity = Δy / Δt.
Substituting t1 = 2 into the equation:
y1 = 50(2) - 16(2)^2
y1 = 100 - 16(4)
y1 = 100 - 64
y1 = 36
Substituting t2 = 2.5 into the equation:
y2 = 50(2.5) - 16(2.5)^2
y2 = 125 - 16(6.25)
y2 = 125 - 100
y2 = 25
Now, calculate the change in height:
Δy = y2 - y1 = 25 - 36 = -11 ft
Calculate the change in time:
Δt = t2 - t1 = 2.5 - 2 = 0.5 seconds
Finally, calculate the average velocity:
average velocity = Δy / Δt = -11 / 0.5 = -22 ft/s
Therefore, the average velocity for the time period of 0.5 seconds is -22 ft/s.
You can follow a similar process to find the average velocity for the time periods of 0.1 seconds, 0.05 seconds, and 0.01 seconds.
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