# Trajectory Formula

The trajectory formula is used to find the trajectory or the flight path of a moving object which is moving under the action of gravity. The term trajectory is used for projectiles or heavenly objects. When a stone is thrown in the air, then the parabola is the correct approximation of the path of the projectile.

Let us understand the trajectory formula using solved examples.

## What is the Trajectory Formula?

A trajectory formula is used to tell the path of the projectile. By using this formula, if we know the initial values of the motion, then the exact path of the projectile can be well predicted even without seeing the actual path of the projectile. The formula for the trajectory is:

\(y = x \tan \theta - \dfrac {gx^2}{2v^2 \cos ^2 \theta}\)

Where,

- θ is the angle of projection from horizontal,
- v is the absolute initial velocity,
- g is the acceleration due to gravity.
- y is the horizontal component,
- x is the vertical component

Let us have a look at a few solved examples to understand the trajectory formula better.

## Examples Using the Trajectory Formula

**Example 1:** If the initial velocity of a stone thrown by a boy is 6 m/sec, and the angle at which the stone is thrown is 60^{∘}. Find the equation of the path of the projectile. Use g = 9.8 m/sec^{2}. Solve this by using the trajectory formula.

**Solution:**

Given, θ = 60^{∘}.

v(initial velocity) = 6m /sec

Using the trajectory formula,

\(y = x \tan \theta - \dfrac {gx^2}{2v^2 \cos ^2 \theta}\)

y = x tan 60 - (9.8)(x^{2})/(2)(6^{2})(cos^{2} 60)

y = x√3 - 0.544x^{2}

**Answer:** Hence the equation of the trajectory of the projectile is y = x√3 - 0.544x^{2}.

**Example 2: **If Trevor hits a ball with his bat at an initial velocity of 45 m/s in the air. In the ball's direction of travel, the end of the field is 140.0 m away. If the initial angle at which the ball is thrown is 66.4°. Calculate the vertical height when the ball reaches the end of the field. Solve this by using the trajectory formula.

**Solution:**

Given, θ = 66.4°

v = 45 m/s

x = 140.0 m

Using the trajectory formula,

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

y = 320.6 – 192080/648

y = 320.6 – 296.4 = 24.2

**Answer: **Vertical height when the ball reaches the end of the field is 24.4 m.