DYNAMICS - Lecture Notes / Mehmet Zor

23 Agust, 2024

1.2.2 Projectile Motion (oblique and horizontal throwing motion)

What we want to find:

• When a cannonball (projectile) is fired from the starting point A at a certain angle, we see that its trajectory is curvilinear. We call this projectile motion.

• Although it is seen as a 3-dimensional movement according to the X, Y, Z axis set, in fact, projectile motion (or more generaly terms, oblique shooting motion ) is actually a movement in a plane. The object always moves in a plane (x-y plane in the figure) and the problem is examined in this plane.

• In projectile motion, there is only the gravitational acceleration in the vertical direction.

Known Values

• In projectile problems, the effect of wind is neglected.

(shooting point)

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Figure 1.2.2

Figure 1.2.3

(tvid-1.2.1)

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Projectile Motion is the combination of two linear movements in the x-y plane, which is the plane of motion.

2. on the y direction, rectilinear motion with

constant acceleration

(1.14)

(1.15)

(1.16)

(1.17)

Figure 1.2.4

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

Example 1.2.1

A man on the edge of a 719m slope shoots an arrow at a velocity of 100m/s at an angle of 30^o. Calculate,

a-) the maximum height the arrow will reach,

b-) the velocity of the arrow as it passes the line of where it was first shot and the distance between the arrow and the man,

c-) Calculate the velocity at which the arrow hits the ground and the linear distance between it and the man.

719m

1m

30^o

y

Solution on next page..>>

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Figure 1.2.5

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

Solution:

719m

1m

30^o

D

B

y

x

x_B

h_max

C

At peak D, the velocity is in the horizontal direction :

b-)

O

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Figure 1.2.6

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

719m

1m

30^o

D

B

y

x

C

O

x_C

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Figure 1.2.7

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

Example 1.2.2

Ercan, who is in a balloon that starts to rise upwards at a constant velocity of 2m/s, wants to shoot an arrow horizontally relative to the balloon and hit a target that is 240 m away. Ercan has previously measured that the arrow is flying at a velocity of 80m/s. According to this, a-) Calculate how many seconds after the balloon starts moving, Ercan must shoot the arrow in order for the arrow to hit the target. b-) Find the relative velocity of the arrow relative to the balloon at the moment it hits the target. Take the acceleration due to gravity as g = 10m/s². Neglect the dimensions of the target.

1m

240m

A

D

C

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Ercan

Figure 1.2.8

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

Solution

1m

x=240m

Motion of the balloon → motion with constant veolcity:

A

D

C

b-)

When the target is hit :

(velocity of the arrow)

(velocity of the balloon)

(relative velocity of the arrow relative to the balloon):

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Ercan

a-)

Figure 1.2.9

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

A fighter plane flying parallel to the ground at a constant velocity of 1000 km/h at an altitude of 1500 m above the ground releases a rocket while passing through point A. The rocket hits the target on the ground. If the thrust of the rocket gives it an acceleration of 0.5g in the horizontal direction, calculate the angle θ between the target line and the horizontal line. (g : gravitational acceleration: 9.81 m/s²)

θ= ?

A

1500m

B

1000 km/h

Solution:

The y-coordinate of the rocket when it hits the target is zero. Calculating the elapsed time:

The horizontal path of the rocket is a rectilinear motion with a constant acceleration of 0.5g due to thrust.

Example 1.2.3

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Figure 1.2.10

Figure 1.2.11

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

Question 1.2.1 (*)

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

The cannon shown in the figure fires a projectile with an initial speed of u = 450 km/h. Find the time it takes for the projectile to reach point B and its height at point B. (The acceleration due to gravity will be taken as g = 9.81 m/s².) Answer: h=345.43m

Figure 1.2.12

DYNAMICS - Lecture Notes / Mehmet Zor

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23 Agust, 2024

A projectile will be launched from an anti-aircraft gun with an initial velocity of 540 m/s and angle θ, at an aircraft flying parallel to the ground at a constant speed of 1000 km/h and at an altitude of 6 km. The firing will be done at the moment when the aircraft comes right over the anti-aircraft gun. (The projectile does not have its own thrust.) Accordingly,

a-) What should be the firing angle θ for the projectile to hit the aircraft with a direct hit?

b-) Find the time elapsed until the projectile hits the aircraft.

Answers: a-) 𝜃=59° ; t= 15.51s

θ

6 km

Question 1.2.2 (*)

1.2.2 Kinematics of Particle / Curvilinear Motion / Projectile Motion

Figure 1.2.12