Mechanics Help

Mickey

16th March 2008 - 11:43 AM

I'm having trouble with these questions. If anyone could help me it would be greatly appreciated.

1. An arrow is shot high into the air. It follows a curved path and then lodges into the trunk of a tree at a height of 12.1m above the ground. The tree is 21.7m from the archer. What is the displacement of the arrow?

2. Salmon, swimming to their spawning grounds, leap over all sorts of obstacles. The unoffical salmon-leap record is an amazing 3.45m height. Assuming the fish took off at an angle of 45 degrees, what was the its speed on emerging from the water?

3. A radar station detects an aeroplane approaching directly from the east. At first observation the range to the plane is 1200m at 40 degrees above the eastern horizon. The aeroplane is tracked for another 123 degrees in the vertical east-west plane, the range at final contact being 2580m. Find the displacement of the aeroplane during the period of observation.

Also if someone could check my answer to:
A captain of a ship points the ship in a direction of 60 degrees N of E and commands the crew to set the engines to full steam ahead. The top speed of the ship in still water is 30km/h. If there is an 8km/h ocean current running in a direction 30 degrees N of E what is the velocity of the ship with respect to land?
I got 37.41km/h

N O M

16th March 2008 - 08:29 PM

1. Ignore how the arrow got there. The problem is merely a right angle triangle. You know the height - 12.1m, you know the length - 21.7m.

2. Work out the speed you would get if you dropped the salmon from 3.45m. Now imagine a right angle triangle with two sides equal to that speed, work out the length of the other side. (OK speed != distance, but it still works)

3. It's only East-West movement, so only two dimensions need to be considered.
Ignore the 40 degrees. You have a triangle with two sides 1200m and 2580m long with an agle of 123 degrees between. Find the length of the third side.

Mickey

17th March 2008 - 08:09 AM

Thanks for your help.

N O M

17th March 2008 - 08:26 AM

No problem. It looked like you weren't having trouble with the Maths so much as the wording of the problems.

Essential Nature

18th March 2008 - 04:19 AM

While being accurate in method to solve the equation I am not sure if the answers given have had enough explanation. So let me add a few more points and see if this helps any more.

1) You need to know that displacement is only the distance of a start point to an end point without regard to the path taken.

http://en.wikipedia.org/wiki/Displacement_%28vector%29

It also seems to be assumed that the person who shot the arrow was lying on the ground. Therefore you are asked to calculate the distance from the point shot at ground level to the resting position in the tree. The height is given and the distance is given and so a right triangle is formed.

2) There is a simple principle you need to remember: when an object is dropped from a particular height, the speed at which it hits the ground is also the speed at which it must be launched vertically to achieve the same height. You might want to try to prove this for yourself.

Now since it is launched at an angle there will be a vertical velocity component and a horizontal velocity component. The trajectory angle is the angle with respect to the ground. The velocity in the vertical component will be the initial velocity times the sin of the trajectory angle. The velocity of the horizontal component will be the initial velocity times the cos of the angle.

Remember the sin is the length of the side the angle points to divided by hypotenuse and cos is the length of the side with the angle divided by hypotenuse. This is one fundamental fact you should keep strongly in mind. The sin and cos factor decomposition is important in vector mathematics for physics. It is used in many problems. You probably know this but I just want to point out its importance.

3) Draw the picture out and remember that displacement is point start to point end. Then you will see that NOM had the correct approach and why.

4) Ship problem

Since no point of land is given it seems that you must assume that the new direction is directly away from the point on land of interest. Then your calculations are correct except maybe a typo 37.14 instead of 37.41.

Mickey

18th March 2008 - 10:22 AM

Thanks also for your explanation Essential Nature. I thought you had to do question one like a projectile motion question but I got stuck on it as I didn't seem to have enough values to put into the equations. It seems simple now I look at it. I didn't know that when an object is dropped it has the same speed as if it was launched vertically to achieve that same height. I'll have a look at that. Oh and yes it was a typo, I did get 37.14 instead of of 37.41.

bm1957

18th March 2008 - 10:38 AM

QUOTE (Mickey+Mar 18 2008, 10:22 AM)

I didn't know that when an object is dropped it has the same speed as if it was launched vertically to achieve that same height. I'll have a look at that.

It might help to consider that (change in) speed is equal to acceleration*time.

The acceleration of a body in freefall on earth (this includes an object that has been thrown upwards) is always g, towards the earth (approximately).

So the speed lost by an object travelling up for x metres must be the same as the speed gained by an object travelling down for x metres. (Note that this does not depend on the mass of the object, so the speed of object a after falling through x metres will be equal to the release speed of object b if it is to reach a height of x metres, etc...)

Essential Nature

19th March 2008 - 03:56 AM

There is also another way to view the velocity of an object launched from the ground which is in terms of energy. The energy of an object in motion is ½ mv**2. The energy due to gravitational potential is mgh where h is the distance traveled in the gravitational field near the surface of the earth and where once again the distance h is small enough to ignore the changing force of gravity due to a changing distance from the center of the earth.

Note **2 mean term squared, superscript does not work in this editor.
Also Vo and Ho are terms for initial velocity at ground and maximum height.

Considering an object that is launched straight up and then falls straight down the reasoning of the problem is as follows:

1) Total energy of the system will be constant
E = ½ mv**2 + mgh = constant

2) The reference point is the surface of the earth
The term h is the height from the surface of the earth
At the surface of the Earth the height is zero. This is the launch point where the object is shot straight up. Since ‘h’ at the surface is zero, the reference energy due to the gravitational potential energy will be zero and all of the energy will be in the motion.
E = ½ mVo**2
Vo is the velocity with which the object is shot straight up. After this point the object is moving in free motion with the only force being that applied by gravity.

3) When the object reaches a height Ho where the object is no longer in motion then velocity will be zero and all energy will be in the gravitational potential. That is all of the initial launch energy will be converted into gravitational potential energy.
E = mgHo

Initial energy was E = ½ mVo**2
The total energy is not changed and all energy is converted into gravitational potential energy.

The height reached when the velocity is zero will be determined by the energy of the gravitational potential will be equal to the initial kinetic energy of the launch velocity.

mgHo = ½ mVo**2

Ho = ½ Vo**2/g

When the object then starts falling back to Earth the gravitational potential energy will be converted back motional or Kinetic energy. When the object falls back to the ground the gravitational energy. Thus the gravitational energy mgHo will be converted back into kinetic energy ½ mv**2.

mgHo = ½ mv**2

which of course is

mgHo = ½ mVo**2

Conservation of energy can be a powerful way to understand and solve problems.

Velocity at an intermediate point is determined by

Etotal = ½ mVo**2 = mgHo = ½ mv**2 + mgh

g(Ho – h) = ½ v**2

For objects launched at an angle you decompose it as follows.
Determine the velocity component straight up in the vertical direction
V-vertical = V0 sin(trajectory angle)
and the velocity component in the horizontal direction
V-horixontal = Vo cos(trajectory angle).

The velocity in the horizontal direction is constant. There are no forces in the horzontal direction ignoring wind resistance that is.
The velocity in the vertical direction will be that which is determined by gravity which for this problem is the only force acting upon the object.

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