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Admin عدد الرسائل : 1965 تاريخ التسجيل : 22/01/2008

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) = x_o = 0 ; y(t=0) = y_o = 0

v_x(t=0) = v_ox = 350 m/s; v_y(t=0) = v_oy = 0

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

x(t) = (1/2) a_x t² + v_ox t + x_o y(t) = (1/2) a_y t² + v_oy t + y_o

v_x(t) = a_x t + v_ox v_y(t) = a_y t + v_oy

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. Cool

t²

v_x = 350 m/s v_y(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 à 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 37^o 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; v_0x = v₀ cos 37; v_0y = v₀ sin 37

The acceleration is: a_x = 0; a_y = ‑ 32 ft/sec².

03-2

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

x(t) = (1/2) a_x t² + v_ox t + x_o	y(t) = (1/2) a_y t² + v_ox t + y_o
v_x(t) = a_x t + v_ox	v_y(t) = a_y t + v_oy

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

v_y(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' à t' = 0, or t' = 1.8 sec.

Hence the ball will land at x(t') = x(1.8s) = (48)(4/5)(1. Cool

= 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,

v_ave = (x₂ ‑ x₁)/(t₂ ‑ t₁) = (69 ‑ 100)/(1. Cool

= ‑ 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 60^o 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:

x_o = 0 v_ox = v_o cos 60 = (.5) v_o a_x = 0 .

y_o = 0 v_oy = v_o sin 60 = (.866) v_o a_y = ‑ 32 ft/s² .

We note that the quantity v_o 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) v_o t y(t) = ‑ (1/2)(32) t² + (.866)v_o t

v_y(t) = ‑ 32 t + (.866)v_o .

Since v_o is not given in the problem, some other piece of information must be given. We read that the