Suvat Equations Puns

(Apologies if this has been asked loads before btw)

I have a forces test tommorow so I've been doing revision all week, but I still can't seem to remember any of the SUVAT equations, no matter how hard I try

I know its short time now, but any tips on remembering them?

(I originally posted this question 10 days ago but never got a response, which is totally fair, people aren't obligated to answer of course, but I figured I'd try once more.)

I have a question that seems kind of dumb but I wanted to ask anyway. (I'll note I know little about physics other than studying some of the basics, from various sources, in my spare time.)

I've been going over a Sebastian Lague video about the Kinematic/SUVAT equations, and one of the first examples is getting Mario to jump, which I did in Unity. https://youtu.be/v1V3T5BPd7E?t=90

(As an aside, I had to get help on the Unity forum because the character wasn't reaching the right height, which lead into starting reading about explicit and semi-Euler integration and related; still not done reading about it but the forum help and "correction factor" amount to add was much appreciated.)

That went relatively well, but I'm curious about using a SUVAT equation for basic horizontal movement. Is there an equation that is "best" for that? I can basically do "if (haven't reached maxSpeed) rb2d.velocity += acceleration * Time.deltaTime" which seems like the equation "v = u + at". (I guess it could depend on what I wanted to do, like if I wanted to go a certain distance in a time frame or something else a bit more complex than what I currently have; the pseudocode above is just adding acceleration until the maxAccel is reached.)

Thanks for any advice. Obviously this isn't a "serious" issue by far, but I was curious.

Right now I'm using this script to move horizontally (I'm sure it could use more refactoring but this is just testing of course), and the horizontal momentum can't be adjusted in air.

Here's my original horizontal movement testing (basic application, but seems like "v = u + at"), and here's the jumping script with commented out horizontal testing using the same initialVelocity amount that the jumping does (applied basically the same as the original script I think, so probably not as important to include)

This post follows on from a comment by u/JonTheBold about an old Dragon Magazine article which talks about the same thing.

TL;DR fall damage being 1d6 per 10 feet is consistent with physics.

In D&D, when characters fall large distances, the damage is 1d6 for each ten feet fallen to a maximum of 20d6. Ordinarily when justifying this choice there are a couple of good arguments made; the game doesn't need to be a simulation just a set of balanced rules; damage is an abstraction anyway. But when you think about it, it is actually consistent with physics.

Often we feel that fall damage ought to scale non-linearly with height; i.e. the higher you get, the faster the amount of damage increases. But why is this? Most likely it's because velocity scales with the square of the distance fallen. Take the SUVAT equation (equation of motion under uniform acceleration);

v^(2) = u^(2) + 2as (1)

In this instance, we're talking about an object falling from a great height. v is the object's final velocity, the instant it strikes the ground, u is the object's initial velocity, in this case we can say it's 0, a is the acceleration due to gravity (9.81m/s^(2)), and s is the distance fallen. Simplifying;

v^(2) = 2 * 9.81 * s (2)

We can see from this that as s increases linearly (goes up by 10 feet) that v increases quadratically (it goes up by the square of 10 feet) so it's understandable that people might feel that a similar quadratic relationship between fall damage and height might be appropriate.

But it ain't the fall that kills you, it's the stopping at the end. The thing that damages you when it comes to falling is the instant that all of the kinetic energy (literally the energy your body has due to its movement) must be converted into other forms of energy the moment you hit the ground. Some of this is converted into sonic energy (the thud/squelch), some of it is passed into the ground, but the majority of it is likely absorbed into your body, which absorbs it by breaking. This means that our fall damage ought to scale not with velocity, but with kinetic energy. The equation for kinetic energy is;

K.E. = 1/2 * m * v^(2) (3)

As you can see it's our old friend v*^(2)*! The m here represents the mass of the falling object (and is effectively another constant), and v represents the velocity of a moving object. So the Kinetic Energy also scales with the square of the velocity. We can use our expression of the fin

For Kinematic equations with constant acceleration, we use the SUVAT equations. However aren't the a,s,v and u vectors? Why don't we put an arrow on top to represent them as vectors?

1. a pebble is projected vertically upwards with speed 10.5ms^-1 from point A at the top of a cliff.

a) Find the greatest height above A reached by the pebble.

b) The pebble reached the bottom of the cliff 5 seconds after being projected. Calculate the height of the cliff

Thanks in advance

Do you need to remember the SUVAT equations for AQA Combined Science Physics Paper 2?

Thanks

First off here is the exact question we have been given:

>4. A ball is thrown from an upper storey of a building. It is given an initial velocity of 6.0 m s-1 at an angle of 30.0° above the horizontal. It strikes the ground 3.5 s later. Determine:

>i) the height above the ground from which the ball was thrown,

>ii) the horizontal distance from the building to where the ball strikes the ground,

>iii) the time it takes the ball to reach a point 12.0 m below the level of throwing.

For the first part I just want to check that I'm right to use the suvat equation s=ut+1/2at^2 ? I thought this would be the correct equation, however I'm stumbling on the acceleration due to gravity. I would've thought that, seeing as the ball is thrown upwards, 'u' should be positive, and 'a' should be negative (gravity been down), but that returns a negative height for the building. The alternative is that it's not "a x t^2 " but (at)^2 , which does return a positive value, but means the building is nearly 600m tall, which seems a bit much for a 3.5s drop time.

The second part is fine, just find the horizontal component of the velocity and multiply it by time. Don't think I've got any problems there.

As for the 3rd part, I'm kind of at a loss. I tried to work out v at 12m of displacement (so where s=12) using v^2 =u^2 +2as, and then using the value for v in s=1/2(u+v)t but they gave back about 0.52s which seems far too low for it to go up to it's max height and back down to 12m, so at this point I'm basically just asking what the hell I need to do.

First time doing mechanics and just wanted to see if what I was doing was correct please.

1. a body starts from rest with a constant acceleration and in 12 seconds has moved a distance of 144m

a) What is the acceleration ? I set u=0, t=12, s=144 and using equation s = ut + 1/2 at^2 I got an answer of 72m/s^-2

b) How fast was it moving after 12 seconds ? I used v^2 = u^2 + 2as, and using my value of "a" I got v=144m/s

2. Object is dropped from rest at a height of 100m. It expereinces gravity (-9.18m/s^2 ) at what speed does it hit the ground ?

I set u=0, s=100, a=-91.8 Using v^2 = u^2 + 2as, I got v^2 = 1836 (technically -1836), meaning v=42.85m/s

I am not to sure on my answers as I tried using other equations to check my answers and kept getting different ones. I also tried calculating other variables and inputting them into other equations and still kept getting different answers....

Anyways thanks for your time :)

I've been at this for hours trying to make a ballistic calculator that uses a radar to find the launch angle needed for a projectile to meet its target. But, my bullets always overshoot because I don't know what the radar elevation angle is measured in and how to convert this value to a robotic pivot angle.

I've tried multiplying the radar elevation angle by 2pi to convert to radians (using this value to find the x and y distances for the function), and then dividing by 2pi for the robotic pivot angle, but I think this is wrong. Or maybe the function I made is wrong?

{ Θ =Tan^(-1) ((-x + sqrt(y^(2) - 4*(-4.9x^(2) / u^(2)) * (-y - (4.9x^(2) / u^(2)))) / (2*(-4.9x^(2) / u^(2)))) }

where, u is the initial velocity, x is the horizontal displacement, and y is the vertical displacement. Derived from SUVAT equations, { y = u*t + 0.5*a*t^(2) }.

Any help would be great

I'm trying to remake an old flash game using Unity. The main mechanic of the game is to make the basketball pass over the mouse.

Here's a video of the game: https://www.youtube.com/watch?v=6sArMC1l_Sc

I've been trying to get the hang of this mechanic for two whole days and I still haven't been able to reproduce it. This is what I've been able to make so far. (forgive me for the colors, my screen gives weird colors when recording it)

[[progress in unity][1]][1]

As you will see, I managed to calculate the trajectory of the ball, and calculate the speed to follow that trajectory. What bothers me is that the vertex of the drawn parabola is not placed on top of the mouse, but it is displaced to the left. The further left the mouse is from the ball, the more this delta between the vertex and the mouse is pronounced.

I think this is happening because I calculate the initial velocity U using only the angle α and the maximum height the ball will reach hmax. I have not found a formula that allows me to "force" the (x,y) point of the mouse as the vertex of the parabola and gives me a way to calculate the velocity I need to apply to the ball for it to reach that (x,y) a its highest peak.

I have written a list the parameters and formulas for the SUVAT equations I have found throughout my research on the internet, so I'll list them right below.

    /*
     * SUVAT EQUATIONS
     * S    | horizontal distance traveled
     * U    | initial velocity
     * Ux   | initial horizontal velocity
     * Uy   | initial vertical velocity
     * V    | final velocity
     * A    | standard acceleration due to gravity on the surface of the earth (≈ 9.807 m/s^2)
     * T    | travel time
     * h    | initial height
     * hmax | maximum height
     * α    | release angle relative to horizontal
     * 
     * 
     * When h = 0:
     * 
     * Ux = U * cos(α)
     * Uy = U * sin(α)
     * T = 2 * Uy / A
     * hmax = (v^2 sin^2(α))/(2 * A)
     * hmax = Uy^2 / (2 * A)
     * S = (v^2 sin(2 α))/A (old)
     * S = 2 * Ux * Uy / A
     * 
     * S  = Ux * T
     * Ux = S/T
     * Uy = (hmax/T + (1/2)) * A * T
     * T = (2 v sin(α))/A
     * 
     * When h != 0:
     * 
     * Ux = U * cos(α)
     * Uy = U * sin(α)
     * t = (Uy + sqrt(Uy^2 + 2 * A * h)) / A
     * S = Ux * [Uy + sqrt(Uy^2 + 2 * A * h)] / A
     * hmax = (h + Uy²) / (2 * A)
     * Uy = sqrt( (hmax - h) / (2*A))
     * U = sqrt(V - 2A * hmax) 
    */

Here's a recap of the equations I am

Hi can anyone help?, so basically I have dislexia and im struggling with speed in exams and keeping concentration . I know it is a common problem with dislexic people but I dont want to make excuses or let it put me down. Sometimes I just realize how everyone else in my class is on question 6 before I have even read question 1, and it is a bit disheartning.

I do not struggle with understanding anything (as long as I dont have to learn through reading). Its just that I want to improve my speed and I feel I need to take a different approach to everyone else, but I dont know what that is or how to combat this.

A little bit of background information... I am currently doing maths, physics and biology A levels and I want to get As to get into my first choice university (by the end of the school year).

People say practicing questions helps but I struggle with concentration and get eaily bored or tired. I usualy do half my homework then do the rest when I feel like it (within the deadline) on a different day or something.

Unusually when I have to repeat an automated prosess I dont tend to remember shorcuts and have to start all over from the beginning before I answer the question. Regardless of how many times I do the question. I always have to work from the beginning to follow the logic rather then rely on a common pattern. Like in suvat equations I always have to write out suvat and the numbers clearly next to the letters before knowing what formula to use.

Does anyone have any advice, sorry if that was a bit of a read. I just feel like speed is a key aspect in exams and I was hoping for some advice that could help? Or any coping strategies in general that would make things easier.

Hello, today I'm going to figure out once and for all exactly how hard the Wild Tiger himself can strike. I'll be using his feat of overpowering Ohma's Indestructible for this calculation.

For the sake of making this work, I'll be relying on two very small and reasonable assumptions which are as follows;

1. Ohma is flying back roughly one pace, or 2.5 feet, with most of these punches as Wakatsuki is closing the gap quite easily by simply stepping forwards.
2. Energy in all of fiction functions fundamentally different from the way it does in the real world, allowing me to quantifiably compare this to feats of completely different varieties such as resisting heat or electricity.

Now let's begin.

We know how far Ohma's being launched, but that's useless if we don't have a timeframe. Thankfully this panel shows that he's being driven maybe half a step upwards as we can see from his left foot being around 5 inches off the ground.

The SUVAT equation for relating time, distance and acceleration is s = 1/2(u + u + at)

s is displacement, u is initial velocity, a is acceleration and t is time. Using Earth's acceleration of 9.80665 M/s^2 and our displacement of 5 inches (13cm), this gives us a time of 0.1624 seconds.

2.5 feet is 0.762 metres, so covering that distance in that time means that Ohma's being sent flying at a whopping 4.69 M/s, or almost 3.5 times average walking speed!

The formula for kinetic energy is E = 1/2 V^2 x M, or one half of the velocity squared times mass. So with Ohma's speed and weight of 85kg, this would mean Wakatsuki's launching him with almost 935 joules of energy.

This means that Wakatsuki is capable of striking with almost 60% as much energy as Mike Tyson. That's some serious power, and since I attached a string of numbers to my assumptions you know that it's completely factual.

Join me next week, where we find out just how many ants Yujiro could lift (spoiler: it's between two and seven thousand)

Do your worst!

For context I'm a Refuse Driver (Garbage man) & today I was on food waste. After I'd tipped I was checking the wagon for any defects when I spotted a lone pea balanced on the lifts.

I said "hey look, an escaPEA"

No one near me but it didn't half make me laugh for a good hour or so!

Edit: I can't believe how much this has blown up. Thank you everyone I've had a blast reading through the replies 😂

I have a question that seems kind of dumb but I wanted to ask anyway. (I'll note I know little about physics other than studying some of the basics, from various sources, in my spare time.)

I've been going over a Sebastian Lague video about the Kinematic/SUVAT equations, and one of the first examples is getting Mario to jump, which I did in Unity. https://youtu.be/v1V3T5BPd7E?t=90

(As an aside, I had to get help on the Unity forum because the character wasn't reaching the right height, which lead into starting reading about explicit and semi-Euler integration and related; still not done reading about it but the forum help and "correction factor" amount to add was much appreciated.)

That went relatively well, but I'm curious about using a SUVAT equation for basic horizontal movement. Is there an equation that is "best" for that? I can basically do "if (haven't reached maxSpeed) rb2d.velocity += acceleration * Time.deltaTime" which seems like the equation "v = u + at". (I guess it could depend on what I wanted to do, like if I wanted to go a certain distance in a time frame or something else a bit more complex than what I currently have; right now I'm just adding acceleration until the maxAccel is reached essentially.)

Thanks for any advice. Obviously this isn't a "serious" issue by far, but I was curious.

Here's my original horizontal movement testing (basic application, but seems like "v = u + at"), and here's the jumping script with commented out horizontal testing using the same initialVelocity amount that the jumping does (applied basically the same as the original script I think, so probably not as important to include)

I'm trying to remake an old flash game using Unity. The main mechanic of the game is to make the basketball pass over the mouse.

Here's a video of the game: https://www.youtube.com/watch?v=6sArMC1l_Sc

I've been trying to get the hang of this mechanic for two whole days and I still haven't been able to reproduce it. This is what I've been able to make so far. (forgive me for the colors, my screen gives weird colors when recording it)

[[progress in unity][1]][1]

As you will see, I managed to calculate the trajectory of the ball, and calculate the speed to follow that trajectory. What bothers me is that the vertex of the drawn parabola is not placed on top of the mouse, but it is displaced to the left. The further left the mouse is from the ball, the more this delta between the vertex and the mouse is pronounced.

I think this is happening because I calculate the initial velocity U using only the angle α and the maximum height the ball will reach hmax. I have not found a formula that allows me to "force" the (x,y) point of the mouse as the vertex of the parabola and gives me a way to calculate the velocity I need to apply to the ball for it to reach that (x,y) a its highest peak.

I have written a list the parameters and formulas for the SUVAT equations I have found throughout my research on the internet, so I'll list them right below.

    /*
     * SUVAT EQUATIONS
     * S    | horizontal distance traveled
     * U    | initial velocity
     * Ux   | initial horizontal velocity
     * Uy   | initial vertical velocity
     * V    | final velocity
     * A    | standard acceleration due to gravity on the surface of the earth (≈ 9.807 m/s^2)
     * T    | travel time
     * h    | initial height
     * hmax | maximum height
     * α    | release angle relative to horizontal
     * 
     * 
     * When h = 0:
     * 
     * Ux = U * cos(α)
     * Uy = U * sin(α)
     * T = 2 * Uy / A
     * hmax = (v^2 sin^2(α))/(2 * A)
     * hmax = Uy^2 / (2 * A)
     * S = (v^2 sin(2 α))/A (old)
     * S = 2 * Ux * Uy / A
     * 
     * S  = Ux * T
     * Ux = S/T
     * Uy = (hmax/T + (1/2)) * A * T
     * T = (2 v sin(α))/A
     * 
     * When h != 0:
     * 
     * Ux = U * cos(α)
     * Uy = U * sin(α)
     * t = (Uy + sqrt(Uy^2 + 2 * A * h)) / A
     * S = Ux * [Uy + sqrt(Uy^2 + 2 * A * h)] / A
     * hmax = (h + Uy²) / (2 * A)
     * Uy = sqrt( (hmax - h) / (2*A))
     * U = sqrt(V - 2A * hmax) 
    */

Here's a recap of the equations I am