# Projectile motion

**Initial velocity (v**_{o}**) and the component of initial velocity (v**_{ox}** and v**_{oy}**)**

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.

If the object moves parabolically as in Figure 1 and 3 then the initial velocity in the horizontal direction (v_{ox}) and the initial velocity in the vertical direction (v_{oy}) are calculated using the equation:

v_{ox} = v_{o} cos θ

v_{oy} = v_{o }sin θ

If the object moving parabolic as figure 2 then v_{o }= v_{ox }(v_{oy }= 0)

**Velocity (v**_{x }**and v**_{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 }= v_{ox} = constant (uniform linear motion)

v_{y }= v_{oy} + g t or v_{y}^{2 }= v_{oy}^{2} + 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 t^{2}

**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:

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 (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 }= v_{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

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:__

v_{ox} = 20 m/s, v_{oy} = 0 m/s, h = 5 m, g = 9.8 m/s^{2}

**a) Time in air**

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

__Known:__ v_{oy} = 0 m/s, h = 5 m, g = 9.8 m/s^{2}

__Wanted:__ t

**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:__ v_{ox }= 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**

v_{tx} = v_{ox} = 20 m/s

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.

__Known:__ v_{oy} = 0, g = 9.8 m/s^{2} , t = 1 s

__Wanted:__ v_{ty}

v_{t }= v_{o }+ g t —> vo = 0

v_{t }= g t

v_{t} = (9.8 m/s^{2})(1 s)

v_{t }= 9.8 m/s

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 🙂

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:__ 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.

__Known:__

v_{ox }= v_{o} cos θ = (60)(cos 30) = (60)(0.87) = 52 m/s

v_{oy }= v_{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.

__Known:__

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.)

g = – 9.8 m/s^{2}

__Wanted:__ h

v_{t}^{2} = v_{o}^{2 }+ 2 g h

^{2 }= 30^{2} + 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:__

v_{oy} = 30 m/s (this is the initial speed of the bullet in the vertical direction)

g = – 9.8 m/s^{2}

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

__Wanted:__ t

h = v_{o} t + ½ g t^{2}

0 = (30) t + ½ (-9.8 m/s^{2}) t^{2}

0 = (30) t – 4.9 t^{2}

(30) t = 4.9 t^{2}

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:__

v_{ox }= 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**

v_{tx} = v_{ox} = 52.2 m/s

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.

__Wanted:__ v_{oy }= 30 m/s, g = -9.8 m/s^{2} , t = 6.12 seconds

__Wanted:__ v_{ty}

v_{ty} = v_{oy} + g t

v_{ty} = (30) + (-9.8)(6.12)

v_{ty} = (30) – (60)

v_{ty }= -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:

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

__Solution:__

First, we calculate the vertical component (v_{oy}) and the horizontal component (v_{ox}) of the initial velocity (v_{o}).

v_{ox} = v_{o} cos 30^{o }= (10 m/s)(0.87) = 8.7 m/s

v_{oy} = v_{o }sin 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.

__Known:__

v_{oy} = 5 m/s, h = – 5 m, g = – 9.8 m/s^{2}

__Wanted:__ t

h = v_{o} t + ½ g t^{2}

-5 = 5 t + ½ (-9.8) t^{2}

-5 = 5 t – 4.9 t^{2}

-4.9 t^{2} + 5 t + 5 = 0

Use the quadratic formula:

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} = v_{ox} = v_{x }= 8.7 m/s

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.

__Known:__ v_{oy} = 5 m/s, g = -9.8 m/s^{2} , t = 1.64 seconds

__Wanted:__ v_{ty}

v_{ty} = v_{oy} + g t

v_{ty} = 5 + (-9.8)(1.64)

v_{ty} = 5 – 16

v_{ty} = -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:

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.

__Known:__ t = 1.64 seconds, v_{x} = 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:__ v_{oy} = 5 m/s, v_{ty }= 0 m/s (vertical component of the velocity at the maximum height = 0 m/s), g = -9.8 m/s^{2}.

__Wanted:__ h

v_{ty}^{2} = v_{oy}^{2} + 2 g h

0 m/s = (5 m/s)^{2 }+ 2(-9.8 m/s^{2})(h)

0 m/s = 25 (m/s)^{2} + (-19.6 m/s^{2})(h)

25 (m/s)^{2 }= -19.6 m/s 2 (h)

h = 25 (m/s)^{2 }: -19.6 m/s^{2} = 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.