# Isaac physics

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Hi, anyone got any ideas for part B of this:

https://isaacphysics.org/questions/r...tp_kin_shm_ext

https://isaacphysics.org/questions/r...tp_kin_shm_ext

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#4

Not worked it through, but the minimum speed would correspond to the roof being tangential to the motion at the point of impact with the roof. So it just "grazes" the roof and there is zero velocity perpendicular to the roof at this impact point. The projectile motion should be quadratic (parabolic), so it's just a case of finding the velocity such that the roof line is tangent (single solution) to the quadratic motion at the point of intersection (impact).

I'd guess the discriminant=0 equation will give a relationship between v and theta, which can then be solved for a minimum v.

Will work it through in the morning if necessary, as not done so now.

I'd guess the discriminant=0 equation will give a relationship between v and theta, which can then be solved for a minimum v.

Will work it through in the morning if necessary, as not done so now.

Last edited by mqb2766; 4 weeks ago

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(Original post by

Not worked it through, but the minimum speed would correspond to the roof being tangential to the motion at the point of impact with the roof. So it just "grazes" the roof and there is zero velocity perpendicular to the roof at this impact point. The projectile motion should be quadratic (parabolic), so it's just a case of finding the velocity such that the roof line is tangent (single solution) to the quadratic motion at the point of intersection (impact).

I'd guess the discriminant=0 equation will give a relationship between v and theta, which can then be solved for a minimum v.

Will work it through in the morning if necessary, as not done so now.

**mqb2766**)Not worked it through, but the minimum speed would correspond to the roof being tangential to the motion at the point of impact with the roof. So it just "grazes" the roof and there is zero velocity perpendicular to the roof at this impact point. The projectile motion should be quadratic (parabolic), so it's just a case of finding the velocity such that the roof line is tangent (single solution) to the quadratic motion at the point of intersection (impact).

I'd guess the discriminant=0 equation will give a relationship between v and theta, which can then be solved for a minimum v.

Will work it through in the morning if necessary, as not done so now.

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#6

You must have done gcse questions where you looked at a line being tangent to a quadratic?

That is the basis for the problem. Get the equation of the roof (line) and the x-y parabolic motion (quadratic) and solve for their intersection/tangent. That means the discriminant should be 0 and that gives you an equation relating v and theta, the initial values. The (v, theta) pairs which satisfy it represent the speed at which the parabolic motion Will graze the roof, for that angle theta. Then choose theta to give minimum v.

post where you get up to.

That is the basis for the problem. Get the equation of the roof (line) and the x-y parabolic motion (quadratic) and solve for their intersection/tangent. That means the discriminant should be 0 and that gives you an equation relating v and theta, the initial values. The (v, theta) pairs which satisfy it represent the speed at which the parabolic motion Will graze the roof, for that angle theta. Then choose theta to give minimum v.

post where you get up to.

Last edited by mqb2766; 4 weeks ago

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(Original post by

You must have done gcse questions where you looked at a line being tangent to a quadratic?

That is the basis for the problem. Get the equation of the roof (line) and the x-y parabolic motion (quadratic) and solve for their intersection/tangent. That means the discriminant should be 0 and that gives you an equation relating v and theta, the initial values. The (v, theta) pairs which satisfy it represent the speed at which the parabolic motion Will graze the roof, for that angle theta. Then choose theta to give minimum v.

post where you get up to.

**mqb2766**)You must have done gcse questions where you looked at a line being tangent to a quadratic?

That is the basis for the problem. Get the equation of the roof (line) and the x-y parabolic motion (quadratic) and solve for their intersection/tangent. That means the discriminant should be 0 and that gives you an equation relating v and theta, the initial values. The (v, theta) pairs which satisfy it represent the speed at which the parabolic motion Will graze the roof, for that angle theta. Then choose theta to give minimum v.

post where you get up to.

Are these the 2 equations you're referring to?

Last edited by Adaal; 4 weeks ago

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#8

That's t-y not x-y. There is a simple linear relationship between x and t, so use it and replace t with x.

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(Original post by

That's t-y not x-y. There is a simple linear relationship between x and t, so use it and replace t with x.

**mqb2766**)That's t-y not x-y. There is a simple linear relationship between x and t, so use it and replace t with x.

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#10

(Original post by

would that just be t=x/vcos(theta)

**Adaal**)would that just be t=x/vcos(theta)

I'm going to be intermittent on/off line this afternoon, but will check when I can.

Last edited by mqb2766; 4 weeks ago

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(Original post by

Of course. So get a quadratic between y and x by replacing t.

I'm going to be intermittent on/off line this afternoon, but will check when I can.

**mqb2766**)Of course. So get a quadratic between y and x by replacing t.

I'm going to be intermittent on/off line this afternoon, but will check when I can.

This is what I got up to. Not sure what to do with the x. Thanks for your help

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#12

Not really sure what you did by equating gradients? You lose the position of the wall and only use the orientation. Just sketch it.

You want to find the single point of intersection (tangent) between the wall and the quadratic, for a given theta.

So write the wall as

y = mx+c

and sub into the quadratic x-y parabola to get a quadratic in x only.

The discriminant=0 (tangent point) will give the v-theta for trajectories that graze the wall (no x or y dependence).

Then just reason about which v is the smallest.

You want to find the single point of intersection (tangent) between the wall and the quadratic, for a given theta.

So write the wall as

y = mx+c

and sub into the quadratic x-y parabola to get a quadratic in x only.

The discriminant=0 (tangent point) will give the v-theta for trajectories that graze the wall (no x or y dependence).

Then just reason about which v is the smallest.

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Is this the correct equation for which the discriminant must be 0?

When I equated the gradients, I was saying that for a tangent, the gradient of the curve and the straight line must be the same.

Last edited by Adaal; 4 weeks ago

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#14

(Original post by

Is this the correct equation for which the discriminant must be 0?

When I equated the gradients, I was saying that for a tangent, the gradient of the curve and the straight line must be the same.

**Adaal**)Is this the correct equation for which the discriminant must be 0?

When I equated the gradients, I was saying that for a tangent, the gradient of the curve and the straight line must be the same.

But getting closer.

Last edited by mqb2766; 4 weeks ago

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This could be one of the possible situations? And yes, you're right, I was foolishly assuming it would be thrown from the same position as in part A

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#16

(Original post by

This could be one of the possible situations? And yes, you're right, I was foolishly assuming it would be thrown from the same position as in part A

**Adaal**)This could be one of the possible situations? And yes, you're right, I was foolishly assuming it would be thrown from the same position as in part A

y = x/sqrt(3) + 10

Although you have to make sure x points in the same direction in both cases? Is one negative?

Then what is the discriminant?

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(Original post by

Yes that's the right diagram, so assuming the origin is the throwing point, the wall would be

y = x/sqrt(3) + 10

Although you have to make sure x points in the same direction in both cases? Is one negative?

Then what is the discriminant?

**mqb2766**)Yes that's the right diagram, so assuming the origin is the throwing point, the wall would be

y = x/sqrt(3) + 10

Although you have to make sure x points in the same direction in both cases? Is one negative?

Then what is the discriminant?

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#18

(Original post by

Thank you, finally got there!

**Adaal**)Thank you, finally got there!

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(Original post by

No problem. What was the dis criminant v-theta relationship? I never worked it out.

**mqb2766**)No problem. What was the dis criminant v-theta relationship? I never worked it out.

Here it is. Definitely not a nice function of theta!

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#20

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