# Homework Help: Relating Δx and Δy in projectile motion

1. Apr 6, 2009

### marational

1. Describe the relationship (e.g. directly proportional, inversely proportional, exponentially proportional, etc.) between the following pair of variables. Makes use of formulas in your explanation:
Δx(maximum horizontal range) and Δy(maximum height) (for variable angles at constant v1)

2. Projectile motion formulae:
vx = Δx/Δxt, vy2^2 = vy1^2 + 2gΔy, Δy = vy1Δt + 1/2gΔt^2
Component trigonomentric relations:
vx= vicosθ, viy = visinθ

3. The problem I'm having is in stating the relationship between Δx and Δy. It's clear that in this projectile motion scenario where the initial speed remains the same while the initial angle of projection varies, the Δy(maximum height) value increases when the angle is raised and the Δx(maximum horizontal range) value decreases proportionately. It would seem that the two are inversely proportionate, but I can't seem to come up with a formula/combination of formulae which would prove this.

2. Apr 13, 2009

### marational

So I really need some help with this question and thought I might try clarifying some things.

The project was to create some sort of device to launch a projectile of any type a distance of 2.0m, 4.0m, and 6.0m. The device had to be able to be set at three different speeds and four different angles of initial projection.
http://img4.imageshack.us/img4/3429/04022009672.jpg [Broken]
http://img27.imageshack.us/img27/531/04022009676.jpg [Broken]
My partner and I came up with a design that is basically a horizontal bar attached to two support wooden boards which are firmly attached to the wooden base. A rod is attached to a ring clamp which the bar runs through. The rod is attached on its other end to a hammer which hits the projectile, a golf ball. There is a little stand attached to the front of the base board which has four golf tees, each of which can hold a golf ball, in order to hit it at the required four angles.
The way the device works is that the hammer arm is pulled back to a specific height depending on the horizontal distance required, and when released, can hit the golf ball the specified distance from one of four angles.

In the lab section of this project, we had to measure the maximum horizontal range of the projectile shot, the initial projection angle, the time it took for the entire aerial trip, and the mass of the golf ball.
Using this data, we had to determine our initial speed and the maximum height achieved.

This discussion question asked that I describe the relationship between Δx and Δy, though I'm not sure if it was meant to be answered with specific reference to my own design as many different and distinct designs were presented from our class.
Anyways, the first part of the question was about relating Δx(maximum horizontal range) and Δy(maximum height) for variable v1(initial speeds) at a constant angle.

Here is the solution I came up with.
The horizontal distance is proportionate to the square root of the maximum height that the ball reaches. v1, or the initial velocity has a vertical component and a horizontal component. The horizontal component is determined through vx= vicosθ, and then the horizontal distance can be determined through vx = ∆x/∆t. The latter formula may be substituted into the former and rearranged, resulting in vi = ∆x/cosθ∆t. The vertical component is determined through viy = visinθ, and the maximum height can be determined through ∆y = (vy2^2 - vy1^2)/2g. The value ∆y is the maximum height which means that the vertical velocity at this point is 0. When 0 is inserted into the value for vy2, we are left with the formula ∆y = - vy1^2/2g. This can be rearranged to solve for vy1, as vy1 = √(-2g∆y), which can be inserted into the trigonometric relation viy = vi sinθ, resulting in a formula which can be rearranged as vi = √(-2g∆y)/sinθ. These two resultant formulae may be equated as they are both equal to vi, resulting in ∆x/cosθ∆t = √(-2g∆y)/sinθ. Comparing Δx and Δy, it is apparent that the horizontal displacement is proportional to the square root of the maximum height.

I'm not entirely sure if this is correct though it seems to work. A friend of mine tells me that the two values cannot be related with any proportions.

So the second part of the question is where I'm stuck.

It asks to relate Δx and Δy for variable angles at a constant initial speed.

I thought at first that relating the two through the formula tanθ = Δy/Δx might work, but I soon realized that it works out to them being directly proportional to each other which doesn't seem to make sense.
It seemed to me that the two should be inversely proportional as if you increase the angle of projection, the vertical component of the initial velocity is increased while the horizontal component is decreased. The opposite would be true when the angle is decreased. However I realized that vertical velocity and horizontal velocity don't translate directly into Δx and Δy, particularly because the projectile will hit the ground.
For example, If the ball were to be hit perfectly horizontally, 100% of the initial velocity would be in the horizontal direction, but this would result in a very low horizontal range.
The only solution that I could possibly come up with was to reuse the formula from before, (∆x/cosθ∆t = √(-2g∆y)/sinθ) and simply say that the horizontal displacement is proportional to the square root of the maximum height.

If I must use a formula to prove this, I don't know where to start.

It's possible though that Δx and Δy are simply unrelatable.

Last edited by a moderator: May 4, 2017
3. Apr 13, 2009

### arunbg

Try eliminating the time of flight variable t from the equations, instead of the initial velocity. Keeping t constant serves no purpose here. And your friend is right, there is no simple proportionality relationship between the horizontal range and max. height attained if initial velocity is constant and θ varies. This would mean that range and height are unrelated to θ, which is false. Additionally, when the angle is kept constant and the initial velocity changed, see what happens.

4. Apr 13, 2009

### rl.bhat

For a constant velocity of projection, as the angle of projection increases, the maximum height increases. But the maximum horizontal range, first increases, reaches a maximum value at 45 degrees and then decreases.

5. Apr 20, 2009

### marational

Thanks a lot for the advice. This is the answer that I've come up with.

Δx not directly relatable to Δy (for variable angles at constant v1)
Horizontal range and maximum height are not directly relatable under variable angles at a constant initial speed because each is affected in a different way by varying projection angles.
For a constant projection speed, as the angle of projection increases, the maximum height also increases. As the angle of projection increases, more of the initial speed of the projectile is directed in the vertical component of its velocity. After the initial and extremely brief acceleration of the projectile up to its initial velocity, the only vertical acceleration the projectile undergoes is a deceleration of 9.805 m/s^2 as a result of gravity. The maximum height of the projectile’s motion is the point at which its vertical velocity is 0. It has come to a stop travelling in the upwards direction, and is about to begin travelling downwards, all as a result of gravity’s effect on the projectile. Because the acceleration due to gravity is constant, it is logical to assume then that the greater the initial vertical velocity of the projectile, the longer it will take for its vertical velocity to decelerate to 0. The longer the projectile is in flight, the greater the distance it can travel. So it is then understandable that if the projectile is launched with a greater vertical velocity, it will take longer to slow to a vertical stop and reach its maximum height, and travelling at a greater vertical velocity over this greater period of time, it will reach a greater maximum height. The formula viy = visinθ can be substituted into the formula ∆y = viy∆t + ½g∆t^2, resulting in ∆y = visinθ∆t + ½g∆t^2. The maximum height is seen to be the difference between the initial speed multiplied by the sine of the projection angle i.e. the initial vertical velocity, multiplied by the time of flight until the projectile vertically decelerates to a stop i.e. the potential vertical distance which may be reached, and half of the acceleration due to gravity multiplied by the time of flight up until the maximum height is reached, squared i.e. the vertical distance that is lost as a result of gravity working against the initial velocity of the projectile. From this formula we can see that the greater the value of viy, or visinθ, the greater the maximum height reached. When sine is applied to an angle, the closer that angle is to 90°, the closer the value is to 1, and the close the angle is to 0°, the closer the value is to 0. Because the initial speed is to remain constant and only the angle of projection is varied θ must be maximized to achieve the greatest possible maximum height. An angle close to 0 will result in the projectile travelling a miniscule vertical distance before dropping back down and hitting the ground in a minimal period of time. It is thus evident that the maximum height increases with the angle of projection.
The horizontal range on the other hand first increases as the angle of projection increases, reaches a maximum at 45°, then decreases as the angle of projection continues to increase. The formula vx = ∆x/∆t can be inserted into the formula vx= vicosθ and rearranged as vi = ∆x/(∆tcosθ). The formula vy2^2 = vy1^2 + 2g∆y can be rearranged to vy1 = √(vy2^2- 2g∆y). This can then be inserted into viy = visinθ, resulting in √(vy2^2- 2g∆y) = visinθ, which can be rearranged as vi = √(vy2^2- 2g∆y)/sinθ. These two resultant equations can be equated producing, ∆x/(∆tcosθ) = √(vy2^2- 2g∆y)/sinθ. Solving for ∆x, the equation is rearranged to ∆x = √(vy2^2- 2g∆y)∆tcosθ/sinθ. Using the identity cosθ/sinθ = 1/tanθ, the equation becomes ∆x = √(vy2^2- 2g∆y)∆t/tanθ. The horizontal range is inversely proportional to the tan of the angle of projection. As the angle approaches 90°, tanθ becomes a large number. Because ∆x is inversely proportional to tanθ, the larger the tanθ, the smaller the horizontal range. In order for horizontal range to be maximized, tanθ should be minimized, which occurs as the angle of projection approaches 0. However, this would not result in the greatest horizontal range as the vertical component must be considered as well in the formula tanθ = viy/vx. In order to achieve the greatest horizontal range, there must be a balance in the initial horizontal velocity and the initial vertical velocity. This is because this example of projectile motion occurs on a perfectly level surface, with no height difference between projection and landing. tanθ and viy are directly proportional, so if the angle were to be minimized (close to 0°) and accordingly tanθ minimized, a very small vertical velocity would result. This would lead to the projectile barely travelling vertically, and thus hitting the ground in a short period of time. This short length of time would not allow for a great horizontal distance to be reached, regardless of how fast the projectile was initially horizontally projected. In order to achieve a balance in the two components, an angle of 45° should be used. tan45° = 1, demonstrating a perfect one to one ratio between the horizontal and vertical components of the initial velocity, and neglecting air friction, would result in the greatest horizontal range. This proves that the horizontal range is maximized at 45°, and steadily decreases as the angle of projection approaches either 0° or 90°.
Thus as Δy increases and decreases as the projection angle increases and decreases, and Δx increases as the projection angle increases up until 45°, and then decreases as the projection angle increases, the two values are not directly relatable under these circumstances.

For the first part of the question, I altered my answer using a more suitable equation, and it seems correct.

Δx α Δy (for constant angle at variable v1)
The horizontal distance is directly proportionate to the maximum height that the ball reaches. vi, or the initial velocity has a vertical component and a horizontal component. The horizontal component is determined through vx= vicosθ, and the horizontal distance can be found through the equation vx = ∆x/∆t. The latter formula may be substituted into the former resulting in ∆x/∆t = vicosθ. This can be rearranged to ∆t = ∆x/(vicosθ). The vertical component is determined through viy = visinθ, and the maximum height can be determined through ∆y = (vy2 + viy)∆t/2. This can be rearranged to viy = 2∆y/∆t - vy2. Substituting this into the trigonometric relation for viy, results in 2∆y/∆t - vy2 = visinθ. This can be rearranged to ∆t = 2∆y/( visinθ + vy2). Since both equations equal ∆t, the two can be equated arriving at ∆x/(vicosθ) = 2∆y/( visinθ + vy2). Comparing Δx and Δy, it is apparent that they are directly proportional to each other regardless of the initial speed, when the projectile is shot at a constant angle.