Yes. It's that time again. I'm stuck. :/
I've tried figuring it out for myself, but to no avail, and I was too bogged down with other work to ask the teacher for help.

2. A child throws a ball vertically upwards and catches it again.
a) Taking the intiail velocity to be 10.0ms-1, g to be 9.8ms-2, and ignoring effects of air resistance, calculate:

i) the total time of flight of the ball

I've started off with.

s=10.0 x t + 1/2 x 9.81 x t^2

But if you don't know the starting displacement then you can't figure it out? :/

Time to top of flightpath = time from top of flightpath back down to starting point.

Starting displacement = 0, as it's starting at the origin

Therefore for the upwards path, 0 = 10t + 1/2 x (-9.81) x t^2

Note that the acceleration is actually deceleration, so g is -ve.


Edit:

You've got a quadratic in terms of t, therefore solve to find t, and double it to get the time.


I think that works...

Hmmm.
I'm still confused. How do I caulcate the t from that? :/

I've always preferred using 9.8 as g than 9.81.

9.8 is, after all, 7²/5. You can have a lot of fun with that.

And they do say take g as 9.8.

So, for the sake of argument, we have

-(7²/10)t² + 10t = 0

By the power of Greyskull the quadratic formula we have:

t = (-10 ± 10)/(-7²/5) = 0 or (5x20)/7² = 100/7²

2t = 200/7², or, if you prefer, 2(10/7)². That's about four seconds. Not sure if that sounds right or not.

Sorry. I'm being a mathematician.

Hmmm.
I'm still confused. How do I caulcate the t from that? :/
You have the following quadratic equation:

0 = 10t - 4.905(t)^2

Take the t^2 onto the other side (as it's negative), then cancel down (you have a t on one side and a t^2 on the other) and solve like a normal linear equation.

then cancel down (you have a t on one side and a t^2 on the other) Ow ow ow!

Sorry, that *works*, but it's risky because you're cancelling a variable. Far better to factorise that t out.

t(g/2t - 10) = 0

t = 0 or 20/g, or 100/7². 0 is, admittedly, silly.

2t = 40/g or 200/7²

'ang on just a minute.... my method's wrong, isn't it?

Wouldn't you work out the maximum height reached, then use that in the earlier-mentioned quadratic to find t to the top of the path, then double t? Blinx, do you have the answer for this question, so as I can check it?


Assume the starting displacement = 0, you know u, the initial velocity (10m/s), you know v, the final velocity (0m/s, as it will be stationary momentarily at the top of the path), you don't know t (you don't care about this), you know a (-9.81 m/s/s), and you want to find s. I'll use dots to show multiplication, as it's less confusing than using 'x'.

Therefore, v^2 = u^2 + 2.a.s
0 = (10)^2 - 19.62s
s = 100/19.62

Then sub s into s = u.t + 1/2.a.(t)^2 to find t to the top of the path.

Double t to give total time of flight.


I get an answer of 2.03s - this does seem quite a long time to me, but the ball is going up to 5.1 m high, so it might be a reasonable answer.

Thanks for all your input guys.

I think your answer seems the most correct Thomasp, two seconds seems a reasonable amount of time.

But thinking about it, it requires alot of knowladge of SUVAT to work these out. Rather complex for beginning of A-level. :/

I next have to calculate the maximum height. : /

But thinking about it, it requires alot of knowladge of SUVAT to work these out. Rather complex for beginning of A-level. :/

Or, calculus! Everyone loves using calculus to derive forgotten formulae from first principles.

Edit:

How about, the ball's velocity will change by 20m/s at 9.8m/s/s, so that will take (20m/s)/(9.8m/s/s) = 2.04s. Why not do it that way?

Edit2:

For maximum height I'd say it'll convert all its kinetic energy (m/2*100) to potential energy (m*g*h), so (m/2*100)=(m*g*h), 50=g*h, therefore h=50/g=5.1. 5.1m. I was taught all those SUVAT equations too, but the fact is that if you understand what you're doing there's really no reason to remember them, and if you don't then you won't fix that by memorising formulae.

But thinking about it, it requires alot of knowladge of SUVAT to work these out. Rather complex for beginning of A-level. :/

I'm sure SUVAT was one of the first things I did in AS Physics...

I next have to calculate the maximum height. : /

Already done. Check my working!


Have a read through my working in reply 7 again and see if you understand it - if not, let me know and I'll explain it in more detail.

Or, calculus! Everyone loves using calculus to derive forgotten formulae from first principles.

Edit:

How about, the ball's velocity will change by 20m/s at 9.8m/s/s, so that will take (20m/s)/(9.8m/s/s) = 2.04s. Why not do it that way?

Edit2:

For maximum height I'd say it'll convert all its kinetic energy (m/2*100) to potential energy (m*g*h), so (m/2*100)=(m*g*h), 50=g*h, therefore h=50/g=5.1. 5.1m. I was taught all those SUVAT equations too, but the fact is that if you understand what you're doing there's really no reason to remember them, and if you don't then you won't fix that by memorising formulae.
Yeah. a = dv/dt. Therefore, dt = dv/a. I love calculus :D

And as the velocity coming down is negative, the velocity of the object as it returns to the starting location will be -10m/s.

Therefore, the change in velocity (dv) is 20m/s (as andrew said), so the change in time is 20 / 9.81 = 2.04s.


But Blinx probably hasn't done much in the way of calculus yet :p

Yeah. a = dv/dt. Therefore, dt = dv/a. I love calculus :D

Don't say that near the mathematician! You'll enrage it! That's engineer/physicist calculus!

Edit: hang on, beginning of A-level? What's it doing there? This is pretty simple algebra; it ought to be in GCSE physics. (Sorry, I forget, GCSE physics is full of political mush and no science.)

Hmm.

I'm still confused.

Sorry. I feel like an idiot. :/ Physics was never really my strong point, Chemistry is the one I'm best at. But that's probably the easiest :/

I took triple science, so we begun doing SUVAT at the end of GSCE. But, it is only now we are going in-depth into it, and assesing situations where we actually need to use it.

Right, using my non-calculus explanation:

We want to find out the maximum height it reaches. Therefore, at the maximum height, the speed will be zero (as it's on the point of coming back down). The acceleration is negative gravity, as you are throwing it up, against gravity. Write down everything you know:

s = ??? (we want to find this)
u = 10 m/s
v = 0 m/s (final velocity)
a = -9.81 m/s/s (m s^-2)
t = --- (don't need it for this equation)

Using the equation v^2 = u^2 + 2.a.s (I am using a dot - . - to signify multiply) - this is one of the SUVAT equations that you should already have covered.
Subbing in, we get:

0 = 10^2 + 2 * (-9.81) * s

Rearrange:

19.62s = 100

Solve to find s = 5.1m (to 1dp)


Now, we want to find the time it takes to reach a height of 5.1m. Write down everything that applies to this condition.
s = 5.1m
u = 10 m/s
v = 0 m/s
a = 9.81 m/s/s
t = ???

Now, we use the equation s = u.t + 1/2.a.(t)^2 (just the t is squared) - this is a quadratic in terms of t.

Sub everything in to give:
5.1 = 10t - 4.905t^2

Rearrange to get a quadratic in terms of t:
-4.905t^2 + 10t - 5.1 = 0

Use the quadratic equation http://upload.wikimedia.org/math/3/e/a/3ea647783b5121989cd87ca3bb558916.png to solve.
In this case, x is t, a is -4.905, b is 10 and c is -5.1

Solve and you get one answer (and as Paul will say, two equal roots :p), t = 1.02s

To get the total flight time (up and back down), the conditions for the downward flight are the reverse of the upward flight, so you just double the time to give 2.04s.


OK now? Or have I just made matters worse?

And for the record, I consider physics A LOT easier than chemistry.

Don't say that near the mathematician! You'll enrage it! That's engineer/physicist calculus!

Edit: hang on, beginning of A-level? What's it doing there? This is pretty simple algebra; it ought to be in GCSE physics. (Sorry, I forget, GCSE physics is full of political mush and no science.)
I know :D Engineers love enraging mathematicians with our "poor" maths :p


Edit:

Andrew's two explanations in reply 9 are probably simpler than my SUVAT methods.

Right, using my non-calculus explanation:

We want to find out the maximum height it reaches. Therefore, at the maximum height, the speed will be zero (as it's on the point of coming back down). The acceleration is negative gravity, as you are throwing it up, against gravity. Write down everything you know:

s = ??? (we want to find this)
u = 10 m/s
v = 0 m/s (final velocity)
a = -9.81 m/s/s (m s^-2)
t = --- (don't need it for this equation)

Using the equation v^2 = u^2 + 2.a.s (I am using a dot - . - to signify multiply) - this is one of the SUVAT equations that you should already have covered.
Subbing in, we get:

0 = 10^2 + 2 * (-9.81) * s

Rearrange:

19.62s = 100

Solve to find s = 5.1m (to 1dp)


Now, we want to find the time it takes to reach a height of 5.1m. Write down everything that applies to this condition.
s = 5.1m
u = 10 m/s
v = 0 m/s
a = 9.81 m/s/s
t = ???

Now, we use the equation s = u.t + 1/2.a.(t)^2 (just the t is squared) - this is a quadratic in terms of t.

Sub everything in to give:
5.1 = 10t - 4.905t^2

Rearrange to get a quadratic in terms of t:
-4.905t^2 + 10t - 5.1 = 0

Use the quadratic equation http://upload.wikimedia.org/math/3/e/a/3ea647783b5121989cd87ca3bb558916.png to solve.
In this case, x is t, a is -4.905, b is 10 and c is -5.1

Solve and you get one answer (and as Paul will say, two equal roots), t = 1.02s

To get the total flight time (up and back down), the conditions for the downward flight are the reverse of the upward flight, so you just double the time to give 2.04s.


OK now? Or have I just made matters worse?

And for the record, I consider physics A LOT easier than chemistry.
Brilliant. :) I totally understand it now. Sorry for being so clueless before. It's better that I grasp SUVAT, because I'm sure to be usign a helluva lot of it soon.

I'll have a go at the next three questions. :)

Cheers for all your help guys. :D Hopefully this has been interesting for everybody else as it is for me. Maybe we should revive that Maths topic we had ages ago? SSolving complex equations et al. :)

By the power of Greyskull
With the Power of Grayskull, He-Man is only milliseconds away from launching his oversized football straight into the frictionless air with a velocity of 10ms^-1:
http://img529.imageshack.us/img529/6133/hemanad0.jpg

Brilliant. :) I totally understand it now. Sorry for being so clueless before. It's better that I grasp SUVAT, because I'm sure to be usign a helluva lot of it soon.


Excellent!

I'm doing an aeronautical engineering degrees and still use SUVAT - some of the time - when I'm not using weird double and triple derivative/integral equations.

Right.
Stuck.
Again.

A veichle travelling at 24ms-1 and then decelerates uniformly. It travels 48m during braking before coming rest.
a) Calculate.
i) it's average speed duirng deceleration

Well. Distance/Time. 48m/2s = 24m/s

ii)the time it takes to come to rest whilst breaking
Would I need to use.
t = square root of u x t / a?

Part i:

You don't know it takes two seconds - where does that come from? This question is actually more straightforward than it looks.

The vehicle is travelling at 24 m/s at the start, and 0 m/s at the end, and is decelerating at a constant rate. Therefore, it's a simple average:

(24 + 0) / 2 = 12 m/s


Part 2:

Time = Distance / Speed

Therefore 48m / 24 m/s = 2s.

Part i:

You don't know it takes two seconds - where does that come from? This question is actually more straightforward than it looks.

The vehicle is travelling at 24 m/s at the start, and 0 m/s at the end, and is decelerating at a constant rate. Therefore, it's a simple average:

(24 + 0) / 2 = 12 m/s


Part 2:

Time = Distance / Speed

Therefore 48m / 24 m/s = 2s.

I thought it would have been two seconds, cos if it's 24m in one second. And it travels a distance of 48m. 48/24 = 2 seconds? But ****. I just realised. Your not calucating the time for the first question. It's average speed.

Ahh ****. Cheers :)

Now.
Calculating deacceleration.

Accel = v - u / t. So, 0 - 12/2 = 6?

Does that make sense? : /

Now.
Calculating deacceleration.

Accel = v - u / t. So, 0 - 12/2 = 6?

Does that make sense? : /
Yes it does, apart from 0 - 12 is -12, therefore (0 - 12)/2 = -6.

Remember, deceleration is negative, acceleration is positive.

Part 2:

Time = Distance / Speed

Therefore 48m / 24 m/s = 2s.Hang on, what? That's only if the speed's constant, which it isn't. It's going from 24m/s down to 0m/s.

Let's get the numbers out

s = 48m
u = 24m/s
v = 0m/s
a = Meh.
t = What we want to find.

s = ½(u + v)t ("The trapezium rule")

t = 2s/(u + v) = (2 * 48)/24 = 4s.

------

Technically there are five SUVAT equations, although for some reason they don't want you to know all of them (CONSPIRACY!). They are:

1. v = u + at (The obvious one)
2. s = ½(u + v)t (Trapezium rule on a velocity-time graph)
3. s = ut + ½at² (Small rectangle plus triangle rule on a velocity-time graph)
4. s = vt - ½at² (Big rectangle minus triangle rule on a velocity-time graph; the one they don't usually tell you)
5. v² = u² + 2as (The nasty one you have to derive from the others - the simplest way is to sub 1 into 2 for t)

Each of the five equations misses out one of SUVAT. This allows you to calculate one of the five from any three others.

Yeah.
We've been shown all the SUVAT equations, but only the first three have been really explained to us.

It's interesting stuff none the less. =]

Technically there are five SUVAT equations, although for some reason they don't want you to know all of them (CONSPIRACY!). They are:

v = u + at (The obvious one)
s = ½(u + v)t (Trapezium rule)
s = ut + ½at² (Small rectangle plus triangle rule)
s = vt - ½at² (Big rectangle minus triangle rule, the one they don't usually tell you)
v² = u² + 2as (The nasty one you have to derive from the others)

Oh wow, I never learned numbers 2 and 4 in high school.:p And I must say I've never heard the word SUVAT before, but I think I get what it is.

Because I do enjoy hammering points home (and because there is nothing as graphic as a graph to demonstrate a point graphically):

Hang on, what? That's only if the speed's constant, which it isn't. It's going from 24m/s down to 0m/s.

Let's get the numbers out

s = 48m
u = 24m/s
v = 0m/s
a = Meh.
t = What we want to find.

s = ½(u + v)t ("The trapezium rule")

t = 2s/(u + v) = (2 * 48)/24 = 4s.



Dang, I forgot about that. *Slaps self* I could never remember the trapezium rule one at A-level.


I might go and review my differential equations of motion and see how much I can confuse poor Blinx!

How this forum managed to turn basic physics into a 25-post thread is beyond me.

How this forum managed to turn basic physics into a 25-post thread is beyond me.

An engineer, a physicist and a mathematician walked into a thread...

Don't say that near the mathematician! You'll enrage it! That's engineer/physicist calculus!
Enh. Chances are, someone already did the analysis and swept it for mines.

(Wahey! New metaphor. Analysts: the minesweepers for calculus)

EDIT: Or, as my old A-level Maths teacher put it, "we multiply both sides of the equation through by dt, start laughing, then we put integral signs on both sides and now we can stop laughing".

Edit: hang on, beginning of A-level? What's it doing there? This is pretty simple algebra; it ought to be in GCSE physics. (Sorry, I forget, GCSE physics is full of political mush and no science.)It gets worse, though. I remember doing exponential decays and simple harmonic motion in A-Level physics. It was all "Pay no attention to the second derivative behind the curtain!". Because, y'know, not everyone was doing Maths. Shocking, I know.

An engineer, a physicist and a mathematician walked into a thread...
Hahaha! :D

An engineer, a physicist and a mathematician walked into a thread...


*Dies laughing*

EDIT: Or, as my old A-level Maths teacher put it, "we multiply both sides of the equation through by dt, start laughing, then we put integral signs on both sides and now we can stop laughing".

That's inefficient. And I'm an engineer :p

And, engineers hate inefficiency.

Where's the t button on my calculator? I only have numbers 1-6 and some other buttons with weird shapes on. One of them looks like two circles stacked on top of each other, a bit like a headless snowman. "Wrrraaaarrrggghhh!" Oh wait, if he had no head he wouldn't say that. What does snow says when it has no head? If I was snow I would say "brrrr" a lot, I imagine. Snow is cold. That's physics for you.