പ്രൊജക്റ്റൈൽ ചലനം

പ്രൊജക്റ്റൈൽ ചലനത്തെയും പരിഹാരങ്ങളുള്ള സാമ്പിൾ പ്രശ്നങ്ങളെയും കുറിച്ചുള്ള ലേഖനം

പ്രാരംഭ പ്രവേഗം (v_o) ഉം പ്രാരംഭ പ്രവേഗത്തിന്റെ ഘടകവും (v_ox ഒപ്പം വി_oy)

പരാബോളിക് ചലനങ്ങൾ എപ്പോഴും പ്രാരംഭ വേഗതയുള്ള ഒരു വസ്തുവാണ്. പരാബോളിക് ചലനം തിരശ്ചീന, ലംബ ദിശകളിലെ ചലനങ്ങളുടെ സംയോജനമായതിനാൽ, പ്രാരംഭ പ്രവേഗത്തിൽ തിരശ്ചീന, ലംബ ഘടകങ്ങളും ഉണ്ട്.

ചിത്രം 1 ലും 3 ലും ഉള്ളതുപോലെ വസ്തു പരാബോളിക് ആയി ചലിക്കുന്നുവെങ്കിൽ തിരശ്ചീന ദിശയിലുള്ള പ്രാരംഭ പ്രവേഗം (v_ox) ലംബ ദിശയിലുള്ള പ്രാരംഭ പ്രവേഗവും (v_oy) സമവാക്യം ഉപയോഗിച്ച് കണക്കാക്കുന്നു:

v_ox = വി_o cos θ

v_oy = വി_o പാപം θ

If the object moving parabolic as figure 2 then v_o = വി_ox (v_oy = 0)

വേഗത (v_x ഒപ്പം വി_y) and Position (x and y)

The speed in the horizontal and vertical directions at a certain time interval is calculated using the equation:

v_x = വി_ox = constant (ഏകീകൃത രേഖീയ ചലനം)

v_y = വി_oy + g t or v_y² = വി_oy² + 2 g h (free-fall motion)

The position of objects in the horizontal (x) and vertical (y) directions at a certain time interval is calculated using the equation:

x = v_ox t

y = v_oy t + 1⁄2 g t²

Resultant of velocity (v) and position (h)

Resultant velocity at a specified time interval is calculated using the equation:

The direction of objects at a certain time interval is calculated using the equation:

കുറിപ്പുകൾ:

1. The horizontal component of parabolic motion is reviewed as the uniform linear motion, so vox = vx is always constant

2. The vertical component of parabolic motion is reviewed as the free-fall motion, so if the object moving parabolic, such as Figure 1 and 3, the vertical component of the velocity of the object at maximum height is zero (v_y = 0). If you throw a marble upright upwards at a maximum height, the object is rest for a moment (v_y = 0) before turning downward. Therefore, the speed of the object moving the parabolic at maximum height = v_x = വി_ox

3. If the object moves parabolic as shown in Figure 2, the vertical component of parabolic motion is viewed as the free-fall motion. If the object moves parabolically as Figure 1 and 3 then the vertical component of parabolic motion is viewed as the upward vertical motion).

Sample problems:

1. A bullet is fired in a horizontal direction with an initial speed of 20 m/s. If the gun is 5 meters above the ground, determine:

(a) time in air

(b) the maximum height

(c) the horizontal distance

(d) the speed of the bullet when it hit the ground

പരിഹാരം:

Motion in the horizontal direction is analyzed like the uniform linear motion, while motion in the vertical direction is analyzed like the free-fall motion.

അറിയപ്പെടുന്നത്:

v_ox = 20 മീ/സെ, വി_oy = 0 m/s, h = 5 m, g = 9.8 m/s²

a) വായുവിലെ സമയം

The solution is as determining the time interval (t) in the free fall motion.

അറിയപ്പെടുന്നത്: v_oy = 0 m/s, h = 5 m, g = 9.8 m/s²

ആവശ്യമുണ്ട്: t

b) The maximum height

Maximum height = h = 5 meters.

c) Horizontal distance (d)

The solution is like determining the distance on the uniform linear motion

അറിയപ്പെടുന്നത്: v_ox = 20 m/s, t = 1 സെക്കന്റ്

ആവശ്യമുണ്ട്: d

d = v t

d = (20 m/s)(1 s) = 20 m

d) Speed when bullet hits the ground

v_tx = വി_ox = 20മി/സെ

v_ty = ?

First, we calculate the final speed in the vertical direction (vty). The solution is like determining the final speed of the free-fall motion.

അറിയപ്പെടുന്നത്: v_oy = 0, g = 9.8 m/s², ടി = 1 സെ

ആവശ്യമുണ്ട്: v_ty

v_t = വി_o + g t —> vo = 0

v_t = ജിടി

v_t = (9.8 m/s²)(1 s)

v_t = 9.8മി/സെ

The speed of a bullet when it hit the ground:

Direction of bullet:

Because v_tx is in the direction of the positive x-axis (to the right) and v_ty is in the direction of the negative y-axis (downward),

the direction of the bullet when it hits the ground is -26.1^o to the positive x-axis (see figure below).

2. The cannon fired a bullet at a 30^o to the horizontal with a speed of 60 m/s. Determine:

(a) the maximum height

(b) the speed of the bullet at maximum height

(c) time in air

(d) the horizontal distance

(e) the speed of the bullet when it hits the ground. Suppose the ground is flat. 🙂

പരിഹാരം:

Motion in the horizontal direction is analyzed like the uniform linear motion, motion in the vertical direction is analyzed like the upward vertical motion.

അറിയപ്പെടുന്നത്: v_o = 60 m/s, teta = 30^o.

Based on known data, we first calculate the vertical (v_oy) and horizontal (v_ox) components of the initial speed (v_o).

a) The maximum height (h)

The solution is like determining the maximum height on the upward vertical motion.

അറിയപ്പെടുന്നത്:

v_ox = വി_o cos θ = (60)(cos 30) = (60)(0.87) = 52 m/s

v_oy = വി_o sin θ = (60)(sin 30) = (60)(0.5) = 30 m/s

a) Maximum height (h)

The solution is like determining the maximum height on the vertical upward motion.

അറിയപ്പെടുന്നത്:

v_oy = 30 m/s (this is the initial speed of the bullet)

v_ty = 0 m/s (At the maximum height, the vertical velocity of the bullet = 0 m / s. This is the final speed.)

ഗ്രാം = – 9.8 മീ/സെ²

ആവശ്യമുണ്ട്: h

v_t² = വി_o² + 2 ഗ്രാം

0² = 30² + 2 (-9.8) മണിക്കൂർ

0 = 900 – 19.6 h

900 = 19.6 മണിക്കൂർ

h = 900 / 19.6

h = 45.9 മീറ്റർ

The maximum height achieved by the bullet = 45.9 meters.

b) The speed at the maximum height

At the maximum height, the speed in the vertical direction = 0 m/s. At maximum altitude, there is the only speed in the horizontal direction. The speed in the horizontal direction at the maximum height is equal to the initial velocity in the horizontal direction, which is 52.2 m/s. The direction of velocity in the horizontal direction is always constant, that is, in the direction of the positive x-axis (if the motion of an object is described in the diagram above)

c) Time in the air

The solution is like determining the time interval (t) in the discussion of the vertical upward motion.

അറിയപ്പെടുന്നത്:

v_oy = 30 m/s (this is the initial speed of the bullet in the vertical direction)

ഗ്രാം = – 9.8 മീ/സെ²

h = 0 m (when the bullet returns to the ground, the displacement of the bullet in the vertical direction = 0 m)

ആവശ്യമുണ്ട്: t

h = v_o ടി + ½ ജിടി²

0 = (30) t + ½ (-9.8 m/s²) ടി²

0 = (30) t – 4.9 t²

(30) t = 4.9 t²

30 = 4.9 ടൺ

t = 30 / 4.9

t = 6.12 സെക്കൻഡ്

Time in air = 6.12 second

d) Horizontal distance (d)

The solution is like determining the distance (d) on the uniform linear motion.

അറിയപ്പെടുന്നത്:

v_ox = 52.2മി/സെ

t = 6.12 സെക്കൻഡ്

ആവശ്യമുണ്ട്: d

d = v t = (52.2 m/s)(6.12 seconds) = 319.5 m

e) Speed when bullet hits the ground

v_tx = വി_ox = 52.2മി/സെ

v_ty = ?

First, we calculate the final velocity in the vertical direction (v_ty). The solution is to determine the final speed on the vertical upward motion.

ആവശ്യമുണ്ട്: v_oy = 30 m/s, g = -9.8 m/s², t = 6.12 seconds

ആവശ്യമുണ്ട്: v_ty

v_ty = വി_oy + ജിടി

v_ty = (30) + (-9.8)(6.12)

v_ty = (30) – (60)

v_ty = -30 മീ/സെ

A negative sign indicates that the direction of the final velocity is downward. Note that the initial velocity in the vertical direction is equal to the final velocity in the vertical direction.

The speed of the bullet when it hit the ground:

The direction of bullet:

Since v_tx is in the direction of the positive x-axis (to the right) and vty is in the direction of the negative y-axis (downward),

the direction of the bullet’s velocity when it hits the ground is -30^o about the positive x-axis (see figure below).

3. A ball is thrown from the edge of a 50-meter-high building with an initial speed of 10 m/s. If the ball is thrown at 30o about the horizontal, determine:

(a) the time interval the ball reaches the ground

(b) the speed of the ball when it strikes the ground

(c) the horizontal distance that can be reached by the ball is measured from the edge of the building

(d) the maximum height reached by the ball

പരിഹാരം:

First, we calculate the vertical component (v_oy) and the horizontal component (v_ox) of the initial velocity (v_o).

v_ox = വി_o cos 30^o = (10 m/s)(0.87) = 8.7 m/s

v_oy = വി_o പാപം 30^o = (10 m/s)(0.5) = 5 m/s

a) Time interval the ball reaches the ground

The solution is like determining the time interval (t) in a vertical upward motion. The magnitude of the vector whose direction is upward is chosen to be positive, the magnitude of the vector whose direction is downward is chosen to be negative. The ball position where it is thrown selected as a reference point. h is negative because the ground surface is below the reference point, g is negative because the direction of acceleration of gravity is downward.

അറിയപ്പെടുന്നത്:

v_oy = 5 m/s, h = – 5 m, g = – 9.8 m/s²

ആവശ്യമുണ്ട്: t

h = v_o ടി + ½ ജിടി²

-5 = 5 t + ½ (-9.8) t²

-5 = 5 ടൺ – 4.9 ടൺ²

-4.9 ടൺ² + 5 ടൺ + 5 = 0

ക്വാഡ്രാറ്റിക് ഫോർമുല ഉപയോഗിക്കുക:

Time in air = time interval since the ball was thrown to reach the ground = 1.64 seconds.

b) Speed of the ball when strikes the surface of the ground

v_tx = വി_ox = വി_x = 8.7മി/സെ

v_ty = ?

First, we calculate the final speed in the vertical direction (v_ty). The solution is like determining the final speed on the vertical upward motion.

അറിയപ്പെടുന്നത്: v_oy = 5 m/s, g = -9.8 m/s², t = 1.64 seconds

ആവശ്യമുണ്ട്: v_ty

v_ty = വി_oy + ജിടി

v_ty = 5 + (-9.8)(1.64)

v_ty = 5 - 16

v_ty = -11 മീ/സെ

The negative sign indicates that the direction of the final velocity is downward.

The speed of the bullet when it hit the ground:

The direction of the bullet’s speed = the direction of the bullet’s motion when it hit the ground:

The speed of the bullet when it hit the ground:

Because vtx is in the direction of the positive x-axis (to the right) and vty is in the direction of the negative y-axis (downward),

the direction of the bullet when it hits the ground is -52o about the positive x-axis (see figure below).

c) Horizontal distance that can be reached by the ball is measured from the edge of the building

The solution is like determining the distance traveled (d) on the uniform linear motion.

അറിയപ്പെടുന്നത്: t = 1.64 seconds, v_x = 8.7മി/സെ

ആവശ്യമുണ്ട്: d

d = v t = (8.7 m/s)(1.64 s) = 14.3 m

d) The maximum height reached by the ball

അറിയപ്പെടുന്നത്: v_oy = 5 മീ/സെ, വി_ty = 0 m/s (vertical component of the velocity at the maximum height = 0 m/s), g = -9.8 m/s².

ആവശ്യമുണ്ട്: h

v_ty² = വി_oy² + 2 ഗ്രാം

0 m/s = (5 m/s)² + 2(-9.8 m/s²)(h)

0 m/s = 25 (m/s)² + (-19.6 m/s²)(h)

25 (m/s)² = -19.6 m/s 2 (h)

h = 25 (m/s)² : -19.6 m/s² = 1.3 മീറ്റർ

The maximum height reached by the ball = 1.3 meters above the top of the building = 1.3 m + 50 m = 51.3 meters above the surface of the ground.