## Derivation of Equation of Trajectory of a Projectile Motion

# Introduction to Projectile Motion

Before moving on to the equation of trajectory of a projectile motion, it is necessary to first understand the basics of projectile motion. We can define what a projectile means. Any object launched into space with only gravity acting on it is referred to as a projectile. Gravity is the main force affecting a projectile. This doesn’t imply that other forces don’t affect it; it merely means that their impact is far smaller than that of gravity. A projectile’s trajectory is its route after being fired. The projectile is something like a baseball that is batted or hurled.

A projectile usually follows a curved path, or as it is known in physics, a parabolic trajectory. We all must’ve come across the projectile motion at some point in our lives. On the curved path of the projectile, the acceleration due to gravity is constant, and it acts towards the centre of the Earth. We will now look at the way in which projectile motion can be solved.

## Solution of a Projectile Motion

We can solve the projectile by resolving the motion of the projectile into two independent rectilinear motions along the x and y axes, respectively.

**Projectile motion**

Suppose we have a projectile that is projected with an initial velocity u and an angle θ with respect to the x-axis. This angle θ

is known as the angle of the projectile. We can then resolve the velocity u into its x and y components as,

ux=ucosθ

uy=usinθ

We can then solve the projectile along the x and y axes using these components.

The three equations of motion for a constant acceleration due to gravity will be used to solve the projectile. These equations are,

The acceleration due to gravity is only along the y-direction and so the velocity along the x-axis will remain constant. We can now find important parameters for projectile motion.

## Time of Flight

The time of flight is the total time that the projectile stays in the air, from the moment that it is projected to the moment it hits the ground. We know that the velocity at the highest point is zero. In our case, the highest point is A and at this point, velocity is zero in the y-direction. We can write the equation of motion for this case as,

This is the time that it takes for the projectile to reach the highest point.

Now the projectile will again take the same amount of time to fall back to the earth. This means that the time of flight (T) can be written as twice the time that it takes for the projectile to reach the highest point. So the flight time will be

## Horizontal Range

As the name suggests, horizontal range is simply the distance that the projectile travels in the horizontal direction. In our case, the horizontal range or simply the range is represented by R. We can calculate the range by using the equation of motion in the x-direction. It was already discussed that the velocity along the x-axis remains constant because no acceleration acts in that direction. Velocity can be written as,

Here, s is the displacement, and t is the time.

The displacement, in our case, will be the range and the time that the projectile stays in flight will be t. We have already calculated the time of flight, and we know the value of v_{x} , which will be the same as u_{x}.

So the range will be,

## Maximum Height of the Projectile

The maximum height of the projectile can be calculated by using the equation of motion in the y-direction. At the maximum height of the projectile, the velocity in the y-direction will be zero. So we can use this equation to find the maximum height H.

## Derivation of the Equation of Trajectory

The projectile follows a parabolic trajectory, or as we have said before, a curved path. We have already derived the Time of Flight, Horizontal Range, and Height of the Projectile. But this does not provide the complete solution for a projectile. We need the equation of trajectory for the complete solution because it will provide the relation between the x and y coordinates at any point of time in the motion of the projectile.

Now we know that the velocity of the projectile in the x-direction is constant throughout the motion. It is,

ux=ucosθ

However, in the y-direction, there is the acceleration due to gravity. The velocity in the y-direction changes with time, so the equation for the velocity in the y-direction can be written as,

We can also write the equation for the displacement in the y-direction as,

Now to relate the x and y directions we can eliminate t from the above equation by using,

Substituting this value of t in the equation for y, we get,

This equation is the equation of trajectory for a projectile. It is also known as the equation of the path of a projectile. This was the whole equation of trajectory derivation. If someone asks to derive an equation of the path of the projectile, this whole derivation will be the answer.

## Conclusion

When a body is launched at a speed that creates an angle with the horizontal, it follows a parabolic trajectory known as a projectile. It can also be defined as any object launched into space with only gravity acting on it. Gravity is the main force affecting a projectile. Other forces are there, but they don’t impact the projectile as much as gravity. The time of flight of a projectile is the time that the projectile stays in flight. The horizontal range is the distance that the projectile covers in the horizontal direction. The time of flight, horizontal range, and maximum height reached by the projectile depends on the initial velocity and the angle of the projectile. The equation of trajectory is given as

## 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:

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 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.

**Formula**

y = x tan θ − gx^{2}/2v^{2}cos^{2 }θwhere,

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/s^{2}.**

**Solution:**

We have,

v = 10, θ = 60

^{o}, x = 4 and g = 9.8Using the trajectory formula we have,

y = x tan θ − gx

^{2}/2v^{2}cos^{2}θ= 4 (tan 60) − (9.8) (4)

^{2}/2(10)^{2}(cos^{2}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/s^{2}.**

**Solution:**

We have,

θ = 30

^{o}, x = 5, y = 9 and g = 9.8Using the trajectory formula we have,

y = x tan θ − gx

^{2}/2v^{2}cos2 θ=> 9 = 5 (tan 30) − (9.8) (5)

^{2}/2v^{2}(cos^{2}30)=> 9 = 2.88 – 4.903(5)² / v

^{2}(1.5)=> v

^{2}= 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/s^{2}.**

**Solution:**

We have,

v = 12, θ = 45

^{o}, y = 15 and g = 9.8Using the trajectory formula we have,

y = x tan θ − gx

^{2}/2v^{2}cos^{2}θ=> 15 = x (tan 45) − (10) x

^{2}/2 (12)^{2}(cos^{2}45)=> 15 = x – 10x

^{2}/144Solve 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/s^{2}.**

**Solution:**

We have,

θ = 30

^{o}, v = 6 and g = 9.8Using the trajectory formula we have,

y = x tan θ − gx

^{2}/2v^{2}cos^{2}θy = x (tan 30) − (9.8) x

^{2}/2(6)^{2}(cos^{2}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/s^{2}.**

**Solution:**

We have,

θ = 60

^{o}, v = 9 and g = 9.8Using the trajectory formula we have,

y = x tan θ − gx

^{2}/2v^{2}cos^{2}θy = x (tan 60) − (9.8) x

^{2}/2(9)^{2}(cos^{2}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/s^{2}.**

**Solution:**

We have,

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

Using the trajectory formula we have,y = x tan θ − gx

^{2}/2v^{2}cos^{2}θ

y = x (tan 45) – (9.8) x^{2}/2(12)^{2}(cos^{2}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 − gx

^{2}/2v^{2}cos^{2}θ

y = x (tan 65) – (9.8) x^{2}/2(8)^{2}(cos^{2}65)

y = 2.14x – 4.903x²/11.43

And as we know, sine over cosine is a definition of a tangent. So, the final trajectory formula may be expressed as:

## Trajectory calculator: how to use

Let’s check out the trajectory of water from a water fountain:

The critical thing to notice is the possible difference in axes scaling, so the angle may not look the way it should on the chart. Remember that in all calculations, air resistance is neglected.

## FAQ

### How do I find the maximum angle in the projectile motion?

To find the angle that maximizes the horizontal distance in the projectile motion, follow the next steps:

### What is the trajectory of a projectile launched at 30° at 10 m/s?

Assuming a launch with **zero initial height**, the trajectory of the projectile follows this equation:

`y = (x × tan(30°)) + (g × x²)/(2 × 10² × cos²(30°))`

### What is the shape of the trajectory of a projectile?

The trajectory followed by a projectile is a **parabola**, hence a quadratic equation in the horizontal coordinate. This motion is a consequence of the action of the force of gravity: a deceleration in the vertical direction transfers a quadratic dependence on the vertical movement.

### How do I calculate the maximum height of a projectile with θ = 40° and v₀=5 m/s?

To calculate it:

- Start from the equation for the vertical motion of the projectile:
**y = vᵧ × t – g × t² / 2**, where**vᵧ**is the initial vertical speed equal to**vᵧ = v₀ × sin(θ) = 5 × sin(40°) = 3.21 m/s**. - Calculate the time required to reach the maximum height: it corresponds to the time at which
**vᵧ = 0**, and it is equal to**t = vᵧ/g = 3.21 / 9.81 = 0.327 s**. - Substitute the values in the equation:

**y = 3.21 × 0.327 – 9.81 × (0.27)² /2 = 0.525 m**.

**Question:** Marshall throws a ball at an angle of 60^{0}. If it moves at the rate of 6m/s and Steve catches it after 4s. Calculate the vertical distance covered by it.

**Solved Examples on Trajectory Formula**

**Example 1**

Firstly, suppose a cricket player hit a ball, guiding it away from the bat at a velocity of 45.0 m/s at an angle of 66.4∘ in relation to the field. Moreover, if the direction of travel of the ball is towards the end of the field which is 140.0 m away. Then what will be the height of the ball when it will reach the end of the field?

**Solution:**

In this question the height of the ball is its vertical position and the horizontal position of attention in this question is end of the field which is x = 140.0 m. Also, we need to solve the vertical position y to know the height of the ball. Moreover, the angle θ and the initial velocity (v0) is present in the question. So, we can use the trajectory formula to solve for y.

**Example 2**

Suppose a water skier plans to set up a stunt in which he is going to jump over a burning obstacle. Also, the fire will be 4.0 m away from the ramp, and it will be 1.0 m taller than the height of the ramp. Moreover, the ramp is inclined at an angle of 36.9∘ in relation to the water. Furthermore, he plans to take off from the ramp at a velocity of 9.0 m/s. So, calculate if he will be able to jump over the flames or not?

**Solution:**

Horizontal position (x) = 4.0 m

Vertical position (y) = ?

Initial velocity )v0) = 9.0 m/s

Lunch angle = 36.9∘

Putting values in the trajectory formula

So, the vertical position of the skier is y ≅ 1.49 m which is more than 1.0 m. Hence, the skier will easily cross the burning obstacle.

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