Question

If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it height in feet after t second is given by y=45t−16t2. Find the average velocity for the time period begining when t=2 and lasting
0.1 seconds=???

Answer 1

Find y when t = 2
Find y when t = 2.1
Take the difference and divide it by 0.1

Answer 2

y1 = y(2)
y2 =y(2.1)
dy/dt = [y(2.1) - y(2)]/ 0.1

@ranga beat me to it.

Answer 3

□(dy/dt)=45-32t
When t=2,
dy/dt=45-32*2=45-64=-19 (negative sign shows the body is coming down)
When t=2.1s
dy/dt=45-32*2 .1=45-67 .2= -22.2
Average velocity =(19+22.2)/2=(41 .2)/2=20.6 feet per sec.but coming down.
Or you can solve like this
Height when t=2 s is
H=45*2-16*2*2=90 -64=26 feet
When t=2.1 s
H=45*2.1-16 *2.1*2.1=94 .5 – 16*4.41 =94.5- 70.56= 23.94
Distance covered in .1 s= 26 - 23.94 =2.06
Average velocity =distance/time=2.06/0.1 =20.6 feet per second.

Answer 4

@ranga @surjithayer @douglaswinslowcooper Hi, I got y(2)=26 and y(2.1)=23.95, got a difference of 2.06 then divided it by 0.1, getting 20.6. But when I entered my answer in, it says it is NOT correct. What did I do wrong?

Answer 5

This is the whole problem:
If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it height in feet after t second is given by y=45t−16t2. Find the average velocity for the time period begining when t=2 and lasting
(i) 01 seconds =

(ii) 001 seconds =

(iii) 0001 seconds =

Finally based on the above results, guess what the instantaneous velocity of the ball is when t=2.

Answer 6

is it 22.2 feet per sec for the first?

Answer 7

@surjithayer apparently not, it says it's incorrect

Answer 8

Is there a decimal missing in (i), (ii) and (iii) above?

Answer 9

Is i) .01 second? (you don't have a decimal which does not make sense because all i) ii) and iii) are the same number 1 without the decimal.

Answer 10

@ranga my bad it's suppose to be
(i) 0.1 seconds =

(ii) 0.01 seconds =

(iii) 0.001 seconds =

Answer 11

For 0.1 seconds what you calculated above (20.6 ft/sec) is correct.
But it should be -20.6 ft/sec (minus sign because the velocity is decreasing as the ball goes up).

Answer 12

velocity negative shows it is coming down.

Answer 13

y = 45t −16t^2
when t = 2, y = 45*2 - 16(2)^2 = 26
i)
when t = 2.1, y = 45*2.1 - 16(2.1)^2 = 23.94
Average velocity = (23.94 - 26) / (2.1 - 1.0) = -2.06 / 0.1 = -20.6 ft/sec

ii) when t = 2.01, find y.

Answer 14

@surjithayer @ranga It worked! Thank you! Just one last question, how do I go about the last question: Finally based on the above results, guess what the instantaneous velocity of the ball is when t=2.

Answer 15

For that you have to first answer iii)

Answer 16

I have:
(i) 0.1 seconds = -20.6

(ii) 0.01 seconds = -19.16

(iii) 0.001 seconds = -19.01

What do they mean instantaneous velocity? How do I work this problem?

Answer 17

Yes. As we take a smaller and smaller time difference, the average velocity will get closer and closer to the instantaneous velocity at t = 2.

We first got -20.6, then -19.16 and then -19.01. So it looks like the value is approaching -19 ft/sec. That should be the guess and it will be the correct answer.

Answer 18

@ranga Wow, thanks for your help Ranga! Your steps and explanations were very clear!

Answer 19

You are welcome.