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Forum index » Science and Technology » Physics
Speed of Arrow after 1 second
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AA
science forum beginner


Joined: 27 Jun 2006
Posts: 3

PostPosted: Tue Jun 27, 2006 12:35 pm    Post subject: Speed of Arrow after 1 second Reply with quote
Question was posed and solved in class but I disagree with answer.

If an arrow is fired directly vertical (Look Out!) at a rate of 90 m/s.
What will its speed be after 1 second. The instructor came up with a
solution of 58.xx m/s during the class.

To me this seems like way to much speed loss after only one second. I would
think it should be around 90m/s - 9.82m/s = 80.18m/s. Ignoring air
resistance.

The instructor used the following formula h(t) = -16t^2 + 90t. I don't know
the basis for this formula.

Any help would be appreciated.

Thanks
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PD
science forum Guru


Joined: 03 May 2005
Posts: 4363

PostPosted: Tue Jun 27, 2006 12:51 pm    Post subject: Re: Speed of Arrow after 1 second Reply with quote
AA wrote:
Quote:
Question was posed and solved in class but I disagree with answer.

If an arrow is fired directly vertical (Look Out!) at a rate of 90 m/s.
What will its speed be after 1 second. The instructor came up with a
solution of 58.xx m/s during the class.

To me this seems like way to much speed loss after only one second. I would
think it should be around 90m/s - 9.82m/s = 80.18m/s. Ignoring air
resistance.

The instructor used the following formula h(t) = -16t^2 + 90t. I don't know
the basis for this formula.

Any help would be appreciated.

Thanks

Wow, what a mess.

If your instructor used that formula to determine the speed after one
second, it's time to go to the department chair and complain that your
instructor doesn't know what the heck he's talking about.

PD
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AA
science forum beginner


Joined: 27 Jun 2006
Posts: 3

PostPosted: Tue Jun 27, 2006 1:05 pm    Post subject: Re: Speed of Arrow after 1 second Reply with quote
In all fairness this is not a physics class, we were studing tangents to a
curve. What is the origin of the formula used in class, and is my line of
thought correct?


"PD" <TheDraperFamily@gmail.com> wrote in message
news:1151412671.021726.88470@y41g2000cwy.googlegroups.com...
Quote:

AA wrote:
Question was posed and solved in class but I disagree with answer.

If an arrow is fired directly vertical (Look Out!) at a rate of 90 m/s.
What will its speed be after 1 second. The instructor came up with a
solution of 58.xx m/s during the class.

To me this seems like way to much speed loss after only one second. I
would
think it should be around 90m/s - 9.82m/s = 80.18m/s. Ignoring air
resistance.

The instructor used the following formula h(t) = -16t^2 + 90t. I don't
know
the basis for this formula.

Any help would be appreciated.

Thanks

Wow, what a mess.

If your instructor used that formula to determine the speed after one
second, it's time to go to the department chair and complain that your
instructor doesn't know what the heck he's talking about.

PD
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N:dlzc D:aol T:com (dlzc)
science forum Guru


Joined: 25 Mar 2005
Posts: 2835

PostPosted: Tue Jun 27, 2006 1:12 pm    Post subject: Re: Speed of Arrow after 1 second Reply with quote
Dear AA:

"AA" <aa@aa.com> wrote in message
news:NC9og.126$u11.54@tornado.ohiordc.rr.com...
Quote:
The instructor used the following formula
h(t) = -16t^2 + 90t. I don't know the basis for this formula.

The local value for g in metric units is 9.81 m/sec^2. It is
evident that the "instructor" chose the value for g from the
Imperial system of units (32.2 ft/sec^2)...

Are you familiar with integration? If not:
velocity = integral( acceleration * dt | t0, t1 )
position = integral( velocity *dt | t0, t1 )
with constant acceleration and initial velocity...
position( t ) = acceleration/2 * t^2 + initial_velocity * t
.... just choose acceleration = -g, since g and initial velocity
are counterdirected.

David A. Smith
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PD
science forum Guru


Joined: 03 May 2005
Posts: 4363

PostPosted: Tue Jun 27, 2006 1:45 pm    Post subject: Re: Speed of Arrow after 1 second Reply with quote
AA wrote:
Quote:
In all fairness this is not a physics class, we were studing tangents to a
curve. What is the origin of the formula used in class, and is my line of
thought correct?

Yes, your line of thought looks correct.

Your instructor was trying to use a formula for the *distance* an arrow
goes, which would be h(t) = 0 + 90(t) - (1/2)(9.Coolt^2
where the first term is the initial height
the second term is the height it would go if it kept going at 90 m/s
the third term is the correction for the changing speed due to the
acceleration of gravity which is 9.8 m/s^2 (and the 1/2 comes from it
being the average of the change in speed over that interval).

Your instructor apparently remembered from his dim recall that the
acceleration is 32 ft/s^2, which produced, with the 1/2, the erroneous
factor of 16.

However, that is not speed, it's distance.

Quote:


"PD" <TheDraperFamily@gmail.com> wrote in message
news:1151412671.021726.88470@y41g2000cwy.googlegroups.com...

AA wrote:
Question was posed and solved in class but I disagree with answer.

If an arrow is fired directly vertical (Look Out!) at a rate of 90 m/s.
What will its speed be after 1 second. The instructor came up with a
solution of 58.xx m/s during the class.

To me this seems like way to much speed loss after only one second. I
would
think it should be around 90m/s - 9.82m/s = 80.18m/s. Ignoring air
resistance.

The instructor used the following formula h(t) = -16t^2 + 90t. I don't
know
the basis for this formula.

Any help would be appreciated.

Thanks

Wow, what a mess.

If your instructor used that formula to determine the speed after one
second, it's time to go to the department chair and complain that your
instructor doesn't know what the heck he's talking about.

PD
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