Trajectory Formula

Trajectory Formula

This is a type of motion in which an object thrown into the air follows a curve due to the action of gravity. Also, a projectile is an object thrown into the air, and a trajectory is the path it follows. Moreover, the curved path of an object was first revealed by Galileo. Also, ballistics is ballistic motion. Also, gravity is a force in the sense that it acts on an object. The Trajectory Formula helps find the gravitational force acting on an object. In addition, trajectories have vertical (y) and horizontal (x) positional components. Assume you launch a projectile at an angle from the horizontal plane with an initial velocity v0. Then, you can find the vertical position of the object from its horizontal position using this formula.

Orbital formulas are used to find the trajectory of a moving object moving under the action of gravity. The term Trajectory Formula is used for projectiles or celestial bodies. A parabola is an exact approximation of a projectile’s Trajectory Formula when a stone is thrown into the air. Solve an example to understand the orbital formula.

What Is The Trajectory Formula?

A ballistic formula is used to specify the trajectory. Using this formula, knowing the initial value of the motion is sufficient to predict the exact path of the projectile without looking at the actual path of the projectile. For more information, students should consult the Extramarks website’s Trajectory Formula.

Examples Using The Trajectory Formula

Example 1: If a boy throws a stone with an initial velocity of 6 m/s and the angle of his throwing stone is 60°. Find the equation for the projectile Trajectory Formula. Use g = 9.8m/sec2. Solve this using the orbital formula.


Give θ = 60∘. v(initial velocity) = 6m/s


y = x tan60 – (9.8)(x2)/(2)(62)(cos 2 60)

y = x√3 – 0.544×2

Answer: So the projectile Trajectory Formula equation is y = x√3 – 0.544×2.

Example 2: Trevor hits a ball with a bat in the air with a muzzle velocity of 45 m/s. The edge of the field is 140.0 m ahead of him in the direction of travel of the ball. If the initial angle of the throw is 66.4°. Calculate the vertical height when the ball reaches the edge of the field. Solve this using the orbital formula.

Ans:Given θ = 66.4°

v = 45 m/s

x = 140.0m


y = (140)(tan 66.4°) – [(9.8)(140)(140)/(2)(45)2(0.4)2]

y = 320.6 – 192080/648

y = 24.2

Answer: The height of the ball when it reaches the edge of the field is 24.4 m.

Example 3: First, suppose a cricketer hits the ball, and with respect to the field, he is at an angle of 66.4° and he carries the ball from the bat with a velocity of 45.0 m/s. Also, if the direction of travel of the ball is toward the edge of the field, it is 140.0m away. How tall will the ball be when it reaches the edge of the field?

Ans:In this question, the height of the ball is its vertical position, and the horizontal position of interest in this question is the edge of the field, x = 140.0 m. We also need to solve for the vertical position y to know the height of the ball. Also included in the question are the angle θ and the initial velocity (v0). Therefore, we can solve for y using the orbital equation.

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