Trajectory Formula

Projectile motion is a type of motion in which an object moves along a bilaterally symmetrical, parabolic direction. The path that the object takes is referred to as its trajectory. A trajectory is the curving route of an item in relation to its speed and gravity. It is a type of motion in which an object launched into the air travels in a curving route under the influence of gravity. It also includes vertical (y) and horizontal (x) position components. The trajectory formula assists us in determining the gravitational force acting on an object. It is used to calculate the trajectory or flight path of a moving object that is subject to gravity’s pull.

Formula

y = x tan θ − gx²/2v² cos² θ

where,

y is the horizontal component,

x is the vertical component,

θ is the angle at which projectile is thrown from the horizontal,

g is a constant called the acceleration due to gravity,

v is the initial velocity of projectile.

Sample Problems

Problem 1. A projectile is thrown at an initial velocity of 10 m/s and an angle of 60^o. Find the horizontal component of the projectile if its vertical component is 4 m. Use g = 9.8 m/s².

Solution:

We have,

v = 10, θ = 60^o, x = 4 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos² θ

= 4 (tan 60) − (9.8) (4)²/2(10)² (cos² 60)

= 1.73 (4) – 4.903 (16/25)

= 3.78 m

Problem 2. A projectile is thrown at an angle of 30^o. Find the initial velocity of the projectile if its horizontal component is 9 m and the vertical component is 5 m. Use g = 9.8 m/s².

Solution:

We have,

θ = 30^o, x = 5, y = 9 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos2 θ

=> 9 = 5 (tan 30) − (9.8) (5)²/2v² (cos² 30)

=> 9 = 2.88 – 4.903(5)² / v² (1.5)

=> v² = 25

=> v = 5 m/s

Problem 3. A projectile is thrown at an angle of 45^o and an initial velocity of 12 m/s. Find the vertical component of the projectile if its horizontal component is 15 m. Use g = 10 m/s².

Solution:

We have,

v = 12, θ = 45^o, y = 15 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos² θ

=> 15 = x (tan 45) − (10) x²/2 (12)² (cos² 45)

=> 15 = x – 10x²/144

Solve the quadratic equation for x.

=> x = 1.175, -1.275

Rejecting the negative value as distance cannot be less than zero, we get

=> x = 1.175 m

Problem 4. A projectile is thrown at an angle of 30^o and initial velocity of 6 m/s. Find the trajectory equation of the projectile. Use g = 9.8 m/s².

Solution:

We have,

θ = 30^o, v = 6 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos² θ

y = x (tan 30) − (9.8) x²/2(6)² (cos² 30)

y = 0.58 x – 4.9(x)²/(72) (1.5)

y = 0.58x – 4.9x²/27

Problem 5. A projectile is thrown at an angle of 60^o and initial velocity of 9 m/s. Find the trajectory equation of the projectile. Use g = 9.8 m/s².

Solution:

We have,

θ = 60^o, v = 9 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos² θ

y = x (tan 60) − (9.8) x²/2(9)² (cos² 60)

y = 1.73 x – 4.9(x)²/(81) (1/4)

y = 1.73x – 4.903x² / 20.25

Problem 6. A projectile is thrown at an angle of 45^o and initial velocity of 12 m/s. Find the trajectory equation of the projectile. Use g = 9.8 m/s².

Solution:

We have,

? = 45^o, v = 12 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos² θ

y = x (tan 45) – (9.8) x²/2(12)² (cos² 45)

y = x – 4.9(x)²/(144) (1/2)

y = x – 4.9x²/72

Problem 7. A projectile is thrown at an angle of 65^o and initial velocity of 8 m/s. Find the trajectory equation of the projectile. Use g = 9.8 m/s2.

Solution:

We have,

? = 65^o, v = 8 and g = 9.8

Using the trajectory formula we have,

y = x tan θ − gx²/2v² cos² θ

y = x (tan 65) – (9.8) x²/2(8)² (cos² 65)

y = 2.14x – 4.903x²/11.43