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Projectiles

TSR George

Can anyone give me a few hints for 12b. and 6?

2016-03-12 16.03.57.jpg413.2KB

2016-03-12 16.04.00.jpg350.7KB

Reply 1

Zacken

For Q6, resolve the velocity horizontally and vertically.

Horizontally:

x=30t

(x is horizontal displacement)

Vertically:

\displaystyle y = 20t - \frac{1}{2}gt^2

(y is vertical displacement)

Path of the projectiles means an equation of the form

y = f(x)

, so you want to eliminate

t

from the second equation.

You know that

t = \frac{x}{30}

, so substitute that into your second equation and that'll give you the path (which will be a quadratic function of

x

, plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).

(edited 8 years ago)

Reply 2

thebrahmabull

12 looks badass. I haven't seen this type before. For part 12a I think the idea is gradient=acceleration =g. But how did you mathematically show it? Mind sharing? :smile:

For part 12B the idea is that they have different constant horizontal speeds I.e. rate of change of distance with time is different for each. I would like to see the solutions if you do them!

Reply 3

Zacken

Original post by thebrahmabull

12 looks badass. I haven't seen this type before. For part 12a I think the idea is gradient=acceleration =g. But how did you mathematically show it? Mind sharing? :smile:

For part 12B the idea is that they have different constant horizontal speeds I.e. rate of change of distance with time is different for each. I would like to see the solutions if you do them!

I'll post up the solutions if the OP doesn't. :smile:

Reply 4

thebrahmabull

Original post by Zacken

I'll post up the solutions if the OP doesn't. :smile:

Thanks

Reply 5

TeeEm

Original post by Zacken

I'll post up the solutions if the OP doesn't. :smile:

we are waiting ...

Reply 6

Zacken

Original post by TeeEm

we are waiting ...

As I am for the OP to reply...

Reply 7

TeeEm

Original post by Zacken

As I am for the OP to reply...

sorry as I misunderstood ... just woke up
I thought you were putting them up.

Reply 8

Zacken

Oh well, since I've done Q12 - might as well put this up, since you're not replying:

\displaystyle \mathbf{r}_{\theta} = (vt \cos \theta)i + (vt \sin \theta - \frac{1}{2}gt^2)j

\displaystyle \mathbf{r}_{\alpha} = (vt \cos \alpha)i + (vt \sin \alpha - \frac{1}{2}gt^2)j

\mathbf{r}_{\theta} - \mathbf{r}_{\alpha} = vt(\cos \theta - \cos \alpha)i + vt(\sin \theta - \sin \alpha)j

The gradient of this line should be fairly easy to find, and it will be in terms of only

\theta

and

\alpha

, which are constants.

Reply 9

TSR George

Original post by Zacken

For Q6, resolve the velocity horizontally and vertically.

Horizontally:

x=20t

(x is horizontal displacement)

Vertically:

\displaystyle y = 30t - \frac{1}{2}gt^2

(y is vertical displacement)

Path of the projectiles means an equation of the form

y = f(x)

, so you want to eliminate

t

from the second equation.

You know that

t = \frac{x}{20}

, so substitute that into your second equation and that'll give you the path (which will be a quadratic function of

x

, plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).

Thank you!

Reply 10

TSR George

Sorry everyone, my wifi stopped working all day :/ Thank you so much for your help!! I can post the answers now @thebrahmabull if you still want them?

Reply 11

TSR George

Original post by Zacken

Oh well, since I've done Q12 - might as well put this up, since you're not replying:

\displaystyle \mathbf{r}_{\theta} = (vt \cos \theta)i + (vt \sin \theta - \frac{1}{2}gt^2)j

\displaystyle \mathbf{r}_{\alpha} = (vt \cos \alpha)i + (vt \sin \alpha - \frac{1}{2}gt^2)j

\mathbf{r}_{\theta} - \mathbf{r}_{\alpha} = vt(\cos \theta - \cos \alpha)i + vt(\sin \theta - \sin \alpha)j

The gradient of this line should be fairly easy to find, and it will be in terms of only

\theta

and

\alpha

, which are constants.

I got the constant gradient part, but how do you find the rate of increase of distance? I think the formula v^2=u^2 + 2as is used because the answer involves a root.

Reply 12

Zacken

Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive?

Reply 13

TSR George

Original post by Zacken

Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive?

Of course! Thank you, I'll give it a go tomorrow :smile:

Reply 14

TSR George

Original post by Zacken

For Q6, resolve the velocity horizontally and vertically.

Horizontally:

x=20t

(x is horizontal displacement)

Vertically:

\displaystyle y = 30t - \frac{1}{2}gt^2

(y is vertical displacement)

Path of the projectiles means an equation of the form

y = f(x)

, so you want to eliminate

t

from the second equation.

You know that

t = \frac{x}{20}

, so substitute that into your second equation and that'll give you the path (which will be a quadratic function of

x

, plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).

Wouldn't t=x/20 ? I got 400y = 600x -4.9x^2 as the final answer, but the book says it's 900y

Reply 15

Zacken

Apologies, memory loss in the short time-span between reading the question from your picture and typing my stuff out, it's been fixed. The

i

direction would be horizontal, so the initial horizontal speed is

30

, so it would be

t = \frac{x}{30}

Reply 16

TSR George

I don't see how the rate of increase should equal u(2(1-cos(theta - alpha))^0.5.
For a) I got (sin theta - sin alpha)/(cos theta - cos alpha)

Reply 17

thebrahmabull

Original post by Zacken

Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive?