# Projectile/ Trajectory equation

• articulate
In summary, The conversation discusses a projectile motion practical where a marble is released from an inclined ramp and leaves a mark on a piece of paper below. The equation given for this motion includes variables such as gravitational acceleration, vertical and horizontal distances, and a constant k which represents air resistance. The conversation goes on to discuss how to derive the equation and provide a simplified version. It then delves into finding the value of k and comparing it with other variables in the equation. Eventually, the conversation concludes with a solution for the equation and provides a summary of the steps taken to reach it.
articulate
Hi guys,

I have some questions about this equation that my class were given before carrying out our practical for projectile motion (what we did was release a marble from the top of an inclined ramp and then when it hits the board which has a piece of carbon paper on it, a mark is left on a piece of paper under the carbon paper..)

Here's the equation:

y/x=[g(1+k^2)x]/2u^2 + k

k is a constant
g=gravitational acceleration
y=vertical distance (between bottom of ramp and top of board)
x=horizontal distance (between plumline and mark on paper)

Questions:

1. What is k really? (I read from some other sites that k is air resistance. Is that true?) From what equation was it derived from?
2. How do I get this equation?

What I've come up with so far..:

y=(u sin Ɵ)t + 1/2 gt^2

x= u cos Ɵ t

y/x= [(u sin Ɵ)t+1/2 gt^2]/(u cos Ɵ t)

simplified it.. and I ended up with

y/x= tan Ɵ + (gt^2)/(2u cos Ɵ t)

So yeah, I'm stuck here.

Not sure whether I'm on the right track..

2. Yes you are on the right track. You already found your k, its only proving the other part of the equation left. You might need some basic trigonometric identities to finish your proving.

How do I continue on? I've tried and I came up with something but I don't think I did it right.

By comparing both equations and using the y=mx+c concept

eq 1: y/x=[g(1+k^2)x]/(2u^2) + k

eq 2: y/x= tan Ɵ + (gt^2)/(2u cos Ɵ t)

So..

tan Ɵ= k

and

[g(1+k^2)]/(2u^2) = (gt^2)/(2u cos Ɵ t)

?

I don't think this is right at all, is it? I shouldn't comparing this:
[g(1+k^2)]/(2u^2) = (gt^2)/(2u cos Ɵ t)

because x is not present in my second equation.

ANYWAY

I just continued on

by comparing this

1+k^2 (from equation 1)

with

t (from equation 2).

Since k= tan Ɵ and t=1+k^2 (not sure this is right at all),

t= 1 + (tan Ɵ)^2

Can someone explain this^ to me? I'm not getting it.

Ok. Never mind. Solved!

Just posting this for anyone who's interested..

So..

x= u cos Ɵ t

Thus,

t=x/(u cos Ɵ)... eq1

y= u sin Ɵ t + (gt^2)/2... eq2

Substitute the t in eq2 with eq1, you'll get this:

y= x tan Ɵ + (x^2 g)/(2 u^2 cos^2 Ɵ)

y/x= tan Ɵ + [(x g)(1 + tan^2 Ɵ)]/ (2 u^2)

y/x= x (g(1 +tan ^2 Ɵ))/2u^2 +k

y/x= x[g(1+k^2)]/2u^2 +k

^^(shown)

tan Ɵ=k

Hi there,

The projectile/trajectory equation you have been given is known as the range equation. It is used to calculate the horizontal distance (x) that a projectile will travel given its initial velocity (u), launch angle (Ɵ), and gravitational acceleration (g). The constant k in this equation represents the effects of air resistance on the projectile's motion.

1. K is indeed air resistance, also known as drag. It is a force that acts in the opposite direction of an object's motion through a fluid (in this case, air). It is derived from the drag equation, which takes into account the density of the fluid, the object's surface area, and its velocity. In the range equation, k is included to account for the decrease in horizontal distance due to air resistance.

2. The range equation can be derived using principles of projectile motion and basic calculus. It is a combination of the equations for horizontal and vertical displacement, and takes into account the effect of air resistance.

Your approach to simplifying the equation is correct. The final form of the range equation is y/x = tan Ɵ + (k/2) * (x/u)^2. This equation can be rearranged to solve for any of the variables (u, Ɵ, x, or y) depending on what information you have.

I hope this helps clarify the range equation and its use in projectile motion. Good luck with your practical!

## 1. What is the projectile/trajectory equation?

The projectile/trajectory equation is a mathematical formula that describes the path of an object in motion under the influence of gravity. It takes into account the initial velocity, angle of launch, and acceleration due to gravity to calculate the position of the object at any given time.

## 2. How is the projectile/trajectory equation derived?

The projectile/trajectory equation is derived from the equations of motion, specifically the equations for displacement, velocity, and acceleration. By combining these equations and taking into account the effects of gravity, the projectile/trajectory equation is obtained.

## 3. What are the variables in the projectile/trajectory equation?

The variables in the projectile/trajectory equation include initial velocity (v0), launch angle (θ), time (t), acceleration due to gravity (g), and the horizontal and vertical components of displacement (x and y). These variables can be manipulated to solve for different aspects of the projectile's motion.

## 4. How is the projectile/trajectory equation used in real-world scenarios?

The projectile/trajectory equation is used in various real-world scenarios, such as in sports like basketball, where the trajectory of a ball can be calculated to make accurate shots. It is also used in military applications to predict the trajectory of projectiles fired from weapons. Additionally, the equation can be used in engineering and physics to study the motion of objects in freefall.

## 5. Is the projectile/trajectory equation affected by external factors?

Yes, the projectile/trajectory equation can be affected by external factors such as air resistance, wind, and variations in gravitational acceleration. These factors can alter the path of the projectile and may need to be taken into account for more accurate calculations.

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