Projectile motion/ Physics help

[froggy]

This is a question on a homework assignment and I just can't figure it out

 

An object a ramp at an angle of 30 degrees from the horizon then lands at the same height after travelling 36m horizontally. What is his his take off speed? (Assume negligible air resistance)

 

So, first of all I’ve tried to calculate the vertical and horizontal components of velocity

 

S = Displacement (m)
U = Initial velocity (ms-¹)
V = Final velocity (ms-¹)
A = Acceleration (ms-²)
T = Time

 

Horizontal

 

S = 36m
U = Vcos30
V = ?
A = 0 (because of no air resistance, or is this wrong?)
T = ?

 

Vertical

 

S = 0
U = Vsin30
V = ?
A = -9.81 (Gravity)
T = ?

 

I’ve got as far as subbing the vertical and horizontal components into s = ut + 1/2 at², giving me

 

36=Vcos30 * t+1/20 * t                  and        0=Vsin30* ½ -9.81 * t²

 

But I’m unsure how I can derive a value for either V or T, no matter how much I rearrange s = ut + ½ at²

 

Can someone point me in the right direction?

 

Edited October 18, 2015 by froggy

[*gentlydoesit]

Personally I'm site at math.. 

Edited October 18, 2015 by *gentlydoesit
because I'm site at math :s

[OlegTinkov]

If you split the projectile in half so that you have a triangle approximation with base 18 and height 18tan30 then you can use V(vertical)=0 as at the top its a turning point for vertical velocity. Using v^2 =u^2 +2as its possible to obtain u(initial vert. comp.)  Boils down to u=sqrt(-2as) 

s=18tan30, a=-9.81m/s/s, v=0m/s, u=_(rearrangement of the above)

initial velocity = u(vertical)/ cos30

 

Its been a while since I'v done that stuff , so it could be complete rubbish.

 

[Joeyeld]

.

 

Edited October 18, 2015 by Joeyeld

[Joeyeld]

.

[TrialsIsHard]

The key to this question is that you know after a time, T the vertical displacement is zero. So working the vertical plane you need to find at which times, T_1 and T_2 the projectile has a vertical displacement of zero, one of the values will be zero obviously. 

After which you can use your non zero value (or expression..) of time into your horizontal component in order to work out v. 

I don't want to just give you the answer, but have a fiddle with these  

The equations of constant acceleration are: 

s=ut+0.5at^2

v=u+at 

v^2 = u^2 +2as 

s=((u+v)/2) t

[froggy]

Thanks for your help

 

So, if I wanted to calculate the vertical component of time, I know that..

 

S = 18tan30
U = ?
V = 0 (final velocity)
A = -9.81ms-²
T = ?

 

Which leads me to the problem that no equation of motion tends for only the 4 variables of displacement, final velocity, acceleration to give time

 

 

 

I’ve got as far as splitting the horizontal component into half to give…

 

 

 

 

 

 

 

 

 

 

 

So from this I should be able to calculate the initial velocity, as I have the same as before

 

S = 18tan30
U = ?
V = 0 (final velocity)
A = -9.81ms-²
T = ?

 

So, using v² = u² + 2as

 

I can rearrange to find

 

U² = 2as – v²

 

∴ u² = 2(9.81)(18tan30) – 0²

 

u² = square roof of 203.9

 

u = 14.3ms-²

 

Although I know this is wrong. Because this velocity only gives a range of 18 meters, and if I was to double the initial velocity it gives a range of more than double 

 

 

EDIT: I could use the value derived for initial velocity in the vertical plane to calculate the T in the vertical. However, the initial velocity is derived and I don't want to carry that error forward, because I'm not happy that that is the correct value for initial velocity. (Or is this initial velocity ONLY in the vertical plane?) 

Edited October 18, 2015 by froggy

[OlegTinkov]

Assuming the approach I suggested is valid, then that value for the initial velocity is indeed only in the vertical direction, hence at that point you want to find the complete velocity (its just the vector resolution) so taking Uvertical and dividing by cos30 will give you the hypotenuse(the complete velocity, which the question asks of) and that would be you finished there.

[manuel]

Where did you get a tan from?

 

[froggy]

Thanks for your reply, although I don't understand how the hypotenuse can be equal to cos30, doesn't cos30 need a coefficient to be a valid hypotenuse? 

[froggy]

Where did you get a tan from?

 

The opposite of the angle(30°) over the adjacent (18m)

[OlegTinkov]

Well it wouldn't be cos30 alone, but 203.9/cos30 (203.9sec30)

[manuel]

using tan makes no sense with relation to that distance.

[OlegTinkov]

I'm sure you can can use v=sqrt(dg/sin2(30)) for a problem like this. 

Edited October 18, 2015 by OlegTinkov
.

[ManxTrialSpaz]

I'm sure you've sorted this

So from this I should be able to calculate the initial velocity, as I have the same as before

 

S = 18tan30
U = ?
V = 0 (final velocity)
A = -9.81ms-²
T = ?

 

So, using v² = u² + 2as

 

I can rearrange to find

 

U² = 2as – v²

 

∴ u² = 2(9.81)(18tan30) – 0²

 

u² = square roof of 203.9

 

u = 14.3ms-²

I haven't read through this but it looks right. From here, finish the velocity triangle. 14.3ms^-1 initial velocity in the vertical; launch V = 14.3/sin(30) = 28.6ms^-1.

 

Then you can check this answer. Work back the time taken for the projectile to reach it's apex, then multiply that with the initial horizontal velocity ( 28.6cos(30) ). This will give you the horizontal travel from launch to apex, which should be half of the total travel (18m in this case). 

[manuel]

No, it's wrong. 18tan30 is completely meaningless 

[froggy]

No, it's wrong. 18tan30 is completely meaningless 

Thanks for your steer, I believe I've cracked this now. The vertical displacement should be 18cos30 = 18sin60

[manuel]

Again, no... You can't apply that to the vertical displacement either ! Unless the guys trajectory is a triangle.... 

 

Edit: unless you're getting 18 from somewhere other than 36/2

Edited October 19, 2015 by manuel

[ManxTrialSpaz]

No, it's wrong. 18tan30 is completely meaningless 

Oh aye, it was the end of a long week.

 

Froggy, you can't map the displacements on a right angled triangle because of the deceleration in the vertical plane. All you can do is map the initial velocities, U, U_x and U_y onto the triangle with a 30° launch angle.

Then you can gather the correct equations for solving, because you know that U_x and U_y are linked, because when S_x = 18, V_y = 0.

Edited October 19, 2015 by ManxTrialSpaz

[manuel]

I'm not sure that helps either. Ignore vertical/horizontal displacement as a means of linking things together. 

things to notice ..,

1. Ooh why 30 degrees? That makes things easier doesn't it?

2. There is only one thing you can say about the horizontal component.

3. Vertically - what goes up must come down.... And, there is only one equation that you can use here too that doesn't use distance (something you don't know and haven't been asked to calculate)

[manuel]

Just as an aside, and you could give this a go - you can get to the vertical height directly but it's much trickier. It's also independent of gravity/initial velocity. 

[ManxTrialSpaz]

I'm not sure that helps either. Ignore vertical/horizontal displacement as a means of linking things together. 

In my mind, the horizontal displacement is an essential constraint for the launch velocity and the most logical number to work with.

 

Sx = Ux*t  ==> t =Sx / Ux ==> t= Sx / Ucos30

Vy=Uy + at = Usin30 + (-9.81)t = Usin30 - 9.81t = 0 ==> U/2 - 9.81t = 0 ==> U = 19.62t

 

U = 19.62*(Sx / Ucos30) ==> U² = 19.62*18/cos30

U=20.19ms^-1

All done using two equations and two minutes of your time?

And the last part is essentially a rearranging of the direct in vacuo equation; Range = V²sin(2*theta)/g.   Using that as a starting point is the only quicker way I see to do this, so I am intrigued how you'd approach this.

[manuel]

In my mind, the horizontal displacement is an essential constraint for the launch velocity and the most logical number to work with.

 

Ok, yes we did things much the same way (my working below g=10 for lazy..)

what I was trying to say was that knowing Sx=18, Vy=0 doesn't tell you anything at all, and doesn't tell you that the initial velocities are linked, I thought it confused matters and suggested that the way to solve would be something to do with how distance varies over time/whatever. 

We were linking together with time - the horizontal distance was a constant.

I wasn't clear either!

 

the reasons I jumped on 18cos30 for max height are A. Pretty sure it's wrong given both our answers of initial velocity at 20ms (intuitively and I worked it out), and B. The way froggy appeared to do the calculation was to pluck height out of thin air and use it to get velocity, which (given I didn't have any of the specific projectile equations) seemed unlikely to be the way to go.

 

at this point if I was going to calculate height I would be using s=ut + .5at^2 as it seems a bit silly to use a specific projectile equation plucked from the air. We know time very simply as we know V. 

 

Interestingly at at this point I thought well actually what if you can pluck the height out without calculating V or t ? The range is fixed at 36m so there is only one path it can possibly follow, for which you don't need to know gravity or initial velocity. You can describe this as a parabola fairly easily. The differential at time 0 and using the angle at take off helps you get the height to drop out. I haven't got this written down legibly! But it gave the same answer which is always good.

 

I will end by saying I haven't done any maths for a long long time so did this as a sanity/mind check, so hey I'm rusty and probably haven't helped.

 

[ManxTrialSpaz]

the reasons I jumped on 18cos30 for max height are A. Pretty sure it's wrong given both our answers of initial velocity at 20ms (intuitively and I worked it out), and B. The way froggy appeared to do the calculation was to pluck height out of thin air and use it to get velocity, which (given I didn't have any of the specific projectile equations) seemed unlikely to be the way to go.

 

I will end by saying I haven't done any maths for a long long time so did this as a sanity/mind check, so hey I'm rusty and probably haven't helped.

 

I must admit, I was too tired to read everything the first couple of times I posted, but froggy was definitely chasing the wrong numbers.

 

I like your approach to the numbers - nice and simple and I can see what you mean by Vy=0 and Sx=18 not really conveying much, but I suppose that's just a shorthand/quick'n'easy way of writing V_y(t/2) = 0 and S_x(t/2) = 18 and obviously the time histories for these are the two equations we've used; V_y(t)=U_y(t) + 0.5a(t)t and S_x(t)=U_x(t)t and that's more accurately what you've done in your working.

I was just cheating and taking a shortcut in the working because we know the trajectory is symmetric.

 

I hope you've got this covered now froggy

[froggy]

Thanks for your all your help guys. much appreciated. I don't know why I approached the question the way I did, I think I just over complicated it.

Here is how I ended up solving it