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  #1  
قديم 06-11-2008, 00:45
maan maan غير متواجد حالياً
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تاريخ التسجيل: Nov 2005
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افتراضي مسائل في حركة المقذوفات وحلولها

03-1
SAMPLE PROBLEMS: 111 SET #4 PROJECTILE MOTION (2 D MOTION PROBLEMS):

1) A bullet has a speed of 350 m/sec as it leaves a rifle. If it is fired horizontally from a cliff 6.4 m above a lake, how far does the bullet travel before striking the water?

Solution: We have a 2 dimensional problem with constant acceleration (acceleration due to gravity). This is a projectile motion problem. The figure is as shown and the coordinate system selected is drawn. The origin is placed at the bullet's location at time t=0. Hence the initial conditions for the problem are:

x(t=0) = xo = 0 ; y(t=0) = yo = 0

vx(t=0) = vox = 350 m/s; vy(t=0) = voy = 0

Since the only force acting is gravity (downward = + y direction), we have: ax = 0; ay = + g = + 9.8 m/sec2. The general solutions for the constant acceleration problem in two dimensions are:

x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + voy t + yo

vx(t) = ax t + vox vy(t) = ay t + voy

Inserting the values of acceleration and the initial conditions gives us the specific equations (applicable to this one particular problem).
x(t) = (350) t y(t) = (1/2)(9.8) t2
vx = 350 m/s vy(t) = 9.8 t

Let t' be the time when the bullet hits the lake. We then know that: y(t') = + 6.4 m. Thus:

y(t') = + 6.4 = + 4.9 t'2  t' = 1.143 sec.

The horizontal (x) position of the bullet at this time is then: x(t') = (350)(1.143) = 400 m.


2) A player kicks a football at an angle of 37o with the horizontal and with an initial speed of 48 ft/sec. A second player standing at a distance of 100 ft from the first in the direction of the kick starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball before it hits the ground?


Solution: We have a projectile motion problem (as far as the football is concerned). Hence we have drawn a figure, chosen a CS, and write down the initial conditions (initial position & velocity) of the football (at t=0).
x0 = 0; y0 = 0; v0x = v0 cos 37; v0y = v0 sin 37

The acceleration is: ax = 0; ay = 32 ft/sec2.
03-2
The general equations of motion for constant acceleration in 2 dimensions are:

x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + vox t + yo
vx(t) = ax t + vox vy(t) = ay t + voy

We insert the known values for acceleration & initial conditions and obtain the specific equations for the football:
x(t) = (48)(4/5) t y(t) = (1/2)(32) t2 + (48)(3/5) t vy(t) = 32 t + (48)(3/5)

We can now answer any question regarding the motion of the football. In particular, we are interested in when it hits the ground (call this t'). We have:

y(t') = 0 = 16 t'2 + (48)(3/5) t'  t' = 0, or t' = 1.8 sec.

Hence the ball will land at x(t') = x(1.8s) = (48)(4/5)(1.8) = 69 ft from the origin.

We can now consider the 2nd player. His initial position (t=0) is 100 ft from the origin, and he must reach a point 69 ft from the origin in 1.8 sec if he is to catch the ball. Thus from the definition of average velocity,

vave = (x2 x1)/(t2 t1) = (69 100)/(1.8) = 17 ft/sec.

The negative sign indicates that he must run toward the origin (negative x direction).


3) A projectile shot at an angle of 60o above the horizontal strikes a building 80 ft away at a point 48 ft above the point of projection. (a) Find the initial velocity, (b) Find the magnitude & direction of the velocity when it strikes the building.

Solution: The wording identifies the problem as a projectile motion problem. We draw a figure, choose a CS, and write down the initial conditions & acceleration in the problem.

As before, since this is a 2 dimensional problem, initial position, initial velocity, and acceleration are specified by two numbers:


xo = 0 vox = vo cos 60 = (.5) vo ax = 0 .

yo = 0 voy = vo sin 60 = (.866) vo ay = 32 ft/s2 .
We note that the quantity vo is not given in the problem. Hence, our first task will be to determine this quantity. Inserting these values into the general equations of motion in 2 dimensions, we have:

x(t) = (.5) vo t y(t) = (1/2)(32) t2 + (.866)vo t

vy(t) = 32 t + (.866)vo .

Since vo is not given in the problem, some other piece of information must be given. We read that the
03-3

projectile: "strikes a building 80 ft away at a point 48 ft above the point of projection". Drawing a figure, we let the instant when the projectile strikes the building be: t'. Then we have:

x(t') = 80 = (.5) vo t'

y(t') = 48 = (1/2)(32)t'2 + (.866)vo t' .



Hence we have 2 equations in 2 unknowns and can solve for both t' and vo. Solving we find: t' = 2.38 seconds; vo = 67.3 ft/sec.

Since we now know vo, then our specific equations of motion are complete, and we can calculate any other quantity associate with the motion. We are specifically asked for the velocity when it strikes the building (at time t = 2.38 sec.). Thus:

vx = vox = (.5)(67.3) = 33.6 ft/sec

vy(t=2.38s) = 32(2.38) + (.866)(67.3) = 17.9 ft/sec .

We then draw the velocity vector from its components calculated above. The magnitude & direction (angle) can then be determined:

v = = 38.1 ft/sec.

tan  = (17.9)/(33.6)  = 28o (below hor. as shown)



4) A basketball player releases the ball 7 ft above the floor when he is 30 ft from the basket. The ball goes through the rim of the basket (which is 10 ft above the floor) 1.5 seconds after release. Find the initial velocity, and the maximum height above the floor reached by the ball.

Solution: The setup for a projectile motion problem is: (1) figure drawn; (2) CS chosen; (3) initial conditions determined; (4) acceleration identified. (Note we have selected a CS with origin (y=0) on floor.) The initial conditions are:

xo = 0 ; vox = vo cos  ; yo = 7 ft ; voy = vo sin  .


The acceleration in the problem is:

ax = 0 ay = 32 ft/s2 .

Inserting these values into the general equations of motion in 2 dimensions, we have:
03-4

x(t) = vo cos  t ; y(t) = (1/2)(32) t2 + (vo sin  ) t + 7 ; vy(t) = 32 t + vo sin  .

At t = 1.5 seconds we have: x(1.5s) = 30 = vox (1.5) y(1.5s) = 10 = 16(1.5)2 + voy(1.5) + 7

Thus: vox = vo cos  = 20 ft/sec; voy = vo sin  = 26 ft/sec.

The magnitude & direction of the initial velocity is then:

vo = = 32.8 ft/sec; tan  = (26)/(20)  = 52.4o

(That is, 52.4o above the horizontal).

The maximum height above the floor occurs at a time t' when vy(t') = 0. Hence:

vy(t') = 0 = 32 t' + 26  t' = .866 sec.

Then y(0.866sec) = 16 (.866)2 + (26)(.866) + 7 = 17.52 ft.


5) You wish to fire an anti-aircraft shell to intercept an enemy plane flying towards you at 600 mph and an altitude of 42,000 ft. If the plane is initially 20 miles away, and your artillery piece has a muzzle velocity of 2000 ft/sec, how long to you have to adjust the piece to an angle of 60o and fire?


Solution: We have a projectile motion problem (as far as the shell is concerned). Hence we have drawn a figure, chosen a CS, have written down the initial conditions (initial position & velocity) of the shell (at t=0).

xo = 0 vox = vo cos 60 = (.5)(2000) = 1000 ft/sec ax = 0 .


yo = 0 voy = vo sin 60 = (.866)(2000) = 1800 ft/sec ay = 32 ft/s2 .

The general equations of motion for constant acceleration in 2 dimensions are:

x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + vox t + yo

vx(t) = ax t + vox vy(t) = ay t + voy

We insert the known values for acceleration & initial conditions and obtain the specific equations for the shell:
x(t) = 1000 t ; y(t) = (1/2)(32) t2 + 1800 t ; vy(t) = 32 t + 1800 .

We can now answer any question regarding the motion of the shell. One question that might occur to us is: Can we hit a plane at 42,000 ft.? Let's determine the maximum height to which the shell will rise.

03-5
vy(t') = 0 = -32 t' + 1800  t' = 56.3 sec.

Then: ymax = y(t') = -(1/2)(32)(56.3)2 + 1800 (56.3) = 50,625 ft.

We can reach the elevation of the plane. The next question is "When will the shell be at the plane's elevation (42,000 ft)?

y(t') = 42,000 = 16 t'2 + 1800 t' > t'2 - 112.5 t'+ 2625 = 0

We have a quadratic equation to solve. Using the quadratic formula we have:

t' = =

This yields the results: t'1 = 33.03 sec; t'2 = 79.47 sec.

At the 1st time the shell will be a distance x(t'=33.03) = 33,030 ft (6.256 mi) from where it was fired. At the 2nd time x(t'= 79.47) = 79,470 ft (15.05 mi).

But the plane travelling at 880 ft/sec will cover a distance of 13.245 mi in 79.47 seconds. Hence, we can't hit the plane with the shell on the way down.

We illustrate in the figure the various distances involved. xh = 6.256 mi is position where we expect to hit the plane; xo = 20 mi is initial position; xf = ?? is position of plane when we fire the shell. Since the plane will travel a distance of (880)(33.03)/(5280) = 5.505 mi while the shell is in the air, then xf = 6.256 + 5.505 = 11.76 mi away.

Now the question is, "How long do we wait till the plane reaches xf? Taking the difference (20 - 11.76) and dividing by the planes speed, gives us 49.44 seconds. There is one other 'minor' question we should ask ourselves. Where will the plane drop its bomb??? If it will release its bomb at a distance greater than the 11.76 miles then we better get the heck out of here!!! A falling bomb will have equations of motion given by: (using CS with y=0 the ground)

xB(t) = 880 t yB(t) = -(1/2)(32) t2 + 42,000 .

Hence, the time to 'fall" is: yB(t') = 0 = -16 t'2 + 42,000  t' = 51.23 sec.

In this time the bomb travels a horizontal distance of (880)(51.23)/5280 = 8.54 mi. Thus the plane plans to release its bomb when it is 8.54 miles from us. But we know, the plane will never reach this point. It will be shot down 11.76 miles away!

رد مع اقتباس
  #2  
قديم 09-11-2008, 21:01
روح الانوثة عبير روح الانوثة عبير غير متواجد حالياً
فيزيائي جـديد
 
تاريخ التسجيل: Nov 2008
المشاركات: 18
افتراضي رد: مسائل في حركة المقذوفات وحلولها

شكرا بس ليش انجليزي
ممكن عربي
.

رد مع اقتباس
  #3  
قديم 02-08-2009, 06:52
عصام حسنين عصام حسنين غير متواجد حالياً
فيزيائي جـديد
 
تاريخ التسجيل: Oct 2007
الدولة: مصر
المشاركات: 4
رد: مسائل في حركة المقذوفات وحلولها

مثال ( 1 ) : رصدت حركة مقذوف فوجد أن أقصى أرتفاع بلغه هو 400 m فوق نقطة القذف وأن مداه
على المستوى الأفقى هو 2400 m ، أوجد سرعة القذف مقداراً واتجاهاً . أوجد أقصى مدى
لقذيفة قذفت بنفس السرعة السابقة على مستوى مائل ظل زاوية ميله مع الأفقى ( 3/4) .

















































مثال ( 2 ) : ينزلق جسيم على مستوى مائل أملس AB
مسافة 8 m بادئا ً من السكون ثم يواصل حركته بعد B
كقذيفة حرة ثم يصطدم بمستوى مائل BC فى النقطة C
أوجد الطول BC وسرعة الوصول للنقطة C والزمن
المنقضى فى الحركة بين C , A .









مثال ( 3 ) : أطلقت قذيفة من نقطة الأصل فوجد أنها تمر بالنقطتين ( 36 , 12 ) , ( 12 , 12 ) والأبعاد
بالأمتار . أوجد سرعة القذف وزاوية القذف والمدى وزمن التحليق .





مسألة ( 1 ) : أطلقت قذيفة أفقيا ً من نقطة A بسرعة
Vo فأصابت هدفاً B كما هو مبين بالشكل ، فإذا أطلقت
من النقطة O بنفس السرعة Vo لتصيب الهدف B
فأوجد زوايا القذف الممكنة ( g = 32 ft / s2 ) .

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