Question

If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after t second is given by y=95t−16t2. Find the average velocity for the time period beginning when t=1 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=1.

Answer 1

To find the average velocity for different time intervals, we need to calculate the displacement of the ball and divide it by the time interval.

Step-by-step explanation:

To find the average velocity for the time period beginning when t=1 and lasting (i) 0.1 seconds, (ii) 0.01 seconds, and (iii) 0.001 seconds, we need to calculate the displacement of the ball during each time interval and then divide it by the corresponding time interval.

(i) For the time interval of 0.1 seconds:

The displacement is given by:

d = y2 - y1

where y2 = 95(1.1) - 16(1.1)^2 and y1 = 95(1) - 16(1)^2

Substituting the values:

d = 104.5 - 79 = 25.5 ft

The average velocity is given by:

v = d / t

:p>where d is the displacement and t is the time interval.

Substituting the values:

v = 25.5 / 0.1 = 255 ft/s

(ii) For the time interval of 0.01 seconds:

The displacement is given by:

d = y2 - y1

where y2 = 95(1.01) - 16(1.01)^2 and y1 = 95(1) - 16(1)^2

Substituting the values:

d = 100.95 - 78.84 = 22.11 ft

The average velocity is given by:

v = d / t

where d is the displacement and t is the time interval.

Substituting the values:

v = 22.11 / 0.01 = 2211 ft/s

(iii) For the time interval of 0.001 seconds:

The displacement is given by:

d = y2 - y1

where y2 = 95(1.001) - 16(1.001)^2 and y1 = 95(1) - 16(1)^2

Substituting the values:

d = 100.595 - 78.596 = 21.999 ft

The average velocity is given by:

v = d / t

where d is the displacement and t is the time interval.

Substituting the values:

v = 21.999 / 0.001 = 21999 ft/s

Based on the above results, the instantaneous velocity of the ball when t=1 will be the same as the average velocity for a very small time interval, which approaches 22000 ft/s.

Answer 2

Instantaneous velocity is
`v=63` at t=1.

Step-by-step explanation:

The height in feet after t second is given by y(t)=95t−16t2.

Average velocity is defined by:

`v_(ave)=(x_(f) - x_(i) )/(t_(f) - t_(i)  )`.

i)

`t_(i)=1\\x_(i)=y(1)=79\\t_(f)=1+0.1\\x_(f)=y(1+0.1)=85.14\\`

⇒
`v_(ave)=61.4`.

ii)

`t_(i)=1\\x_(i)=y(1)=79\\t_(f)=1+0.01\\x_(f)=y(1+0.01)=79.6284\\`

⇒
`v_(ave)=62.84`.

iii)

`t_(i)=1\\x_(i)=y(1)=79\\t_(f)=1+0.001\\x_(f)=y(1+0.001)=79.062984\\`

⇒
`v_(ave)=62.98`.

Instantaneos velocity is defined by:
`v=\lim_(\triangle t \to 0) (\triangle x)/(\triangle t)`

So we have seen "manually" that when
`\triangle t \rightarrow 0`,
`v \rightarrow 63`.

So instantaneous velocity must be
`v=63` at t=1.