Projectile motion

Initial velocity (vo) and the component of initial velocity (vox and voy)

An object which moves parabolic always has an initial speed. Because parabolic motion is a combination of movements in the horizontal and vertical directions, the initial velocity also has horizontal and vertical components.

Projectile motion 1

If the object moves parabolically as in Figure 1 and 3 then the initial velocity in the horizontal direction (vox) and the initial velocity in the vertical direction (voy) are calculated using the equation:

vox = vo cos θ

voy = vo sin θ

If the object moving parabolic as figure 2 then vo = vox (voy = 0)

Velocity (vx and vy) and Position (x and y)

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

vx = vox = constant (uniform linear motion)

vy = voy + g t or vy2 = voy2 + 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 = vox t

y = voy t + 1⁄2 g t2

Resultant of velocity (v) and position (h)

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

Projectile motion 2

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

Projectile motion 3

Notes:

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 (vy = 0). If you throw a marble upright upwards at a maximum height, the object is rest for a moment (vy = 0) before turning downward. Therefore the speed of the object moving the parabolic at maximum height = vx = vox

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 Projectile motion 4

(b) the maximum height

(c) the horizontal distance

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

Solution:

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.

Known:

vox = 20 m/s, voy = 0 m/s, h = 5 m, g = 9.8 m/s2

a) Time in air

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

Known: voy = 0 m/s, h = 5 m, g = 9.8 m/s2

Wanted: t

Projectile motion 5

b) The maximum height

Maximum height = h = 5 meters.

c) Horizontal distance (d)

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

Known: vox = 20 m/s, t = 1 second

Wanted: d

d = v t

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

d) Speed when bullet hits the ground

vtx = vox = 20 m/s

vty = ?

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

Known: voy = 0, g = 9.8 m/s2 , t = 1 s

Wanted: vty

vt = vo + g t —> vo = 0

vt = g t

vt = (9.8 m/s2)(1 s)

vt = 9.8 m/s

The speed of a bullet when it hit the ground:

Projectile motion 6

Direction of bullet:

Projectile motion 7

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 -26.1o to the positive x-axis (see figure below).

Projectile motion 8

2. The cannon fired a bullet at a 30o 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 🙂

Projectile motion 9

Solution:

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

Known: vo = 60 m/s, teta = 30o .

Based on known data, we first calculate the vertical (voy) and horizontal (vox) components of the initial speed (vo).

Projectile motion 10

a) The maximum height (h)

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

Known:

vox = vo cos θ = (60)(cos 30) = (60)(0.87) = 52 m/s

voy = vo 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.

Known:

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

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

g = – 9.8 m/s2

Wanted: h

vt2 = vo2 + 2 g h

2 = 302 + 2 (-9.8) h

= 900 – 19.6 h

900 = 19.6 h

h = 900 / 19.6

h = 45.9 meter

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 as 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.

Known:

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

g = – 9.8 m/s2

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

Wanted: t

h = vo t + ½ g t2

0 = (30) t + ½ (-9.8 m/s2) t2

0 = (30) t – 4.9 t2

(30) t = 4.9 t2

30 = 4.9 t

t = 30 / 4.9

t = 6.12 seconds

Time in air = 6.12 sekon

d) Horizontal distance (d)

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

Known:

vox = 52.2 m/s

t = 6.12 seconds

Wanted: d

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

e) Speed when bullet hits the ground

vtx = vox = 52.2 m/s

vty = ?

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

Wanted: voy = 30 m/s, g = -9.8 m/s2 , t = 6.12 seconds

Wanted: vty

vty = voy + g t

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

vty = (30) – (60)

vty = -30 m/s

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:

Projectile motion 11

The direction of bullet:

Projectile motion 12

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

Projectile motion 13

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

Projectile motion 14

Solution:

First, we calculate the vertical component (voy) and the horizontal component (vox) of the initial velocity (vo).

Projectile motion 15

vox = vo cos 30o = (10 m/s)(0.87) = 8.7 m/s

voy = vo sin 30o = (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.

Known:

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

Wanted: t

h = vo t + ½ g t2

-5 = 5 t + ½ (-9.8) t2

-5 = 5 t – 4.9 t2

-4.9 t2 + 5 t + 5 = 0

Use the quadratic formula:

Projectile motion 16

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

vtx = vox = vx = 8.7 m/s

vty = ?

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

Known: voy = 5 m/s, g = -9.8 m/s2 , t = 1.64 seconds

Wanted: vty

vty = voy + g t

vty = 5 + (-9.8)(1.64)

vty = 5 – 16

vty = -11 m/s

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

The speed of the bullet when it hit the ground:

Projectile motion 17

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).

Projectile motion 18

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.

Known: t = 1.64 seconds, vx = 8.7 m/s

Wanted: d

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

d) The maximum height reached by the ball

Known: voy = 5 m/s, vty = 0 m/s (vertical component of the velocity at the maximum height = 0 m/s), g = -9.8 m/s2.

Wanted: h

vty2 = voy2 + 2 g h

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 meters

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.

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