Slope and velocity problem (difficult)

[IamVector]

Homework Statement

What is the minimum initial velocity that has to be
given to a stone in order to throw it across a sloped roof? The
roof has width b, its two edges have heights a and c.

Relevant Equations

To find a minimum (or a maximum), we have to
vary free parameters (in this case the throwing point and the

angle) by infinitesimally small increments and see what hap-
pens to the quantity of interest. If it increases for all allowed

variations, we have found a minimum.
And
For a free-fall of a body, there is an integral of mo-
tion (quantity which is conserved), 1/2v^2 − gh, where v is an
instantaneous speed, and h is the current height of the body.

(I don't have slightest idea what to do after )
any geometric approach??

 

[PeroK]

Homework Statement:: What is the minimum initial velocity that has to be
given to a stone in order to throw it across a sloped roof? The
roof has width b, its two edges have heights a and c.
Relevant Equations:: To find a minimum (or a maximum), we have to
vary free parameters (in this case the throwing point and the

angle) by infinitesimally small increments and see what hap-
pens to the quantity of interest. If it increases for all allowed

variations, we have found a minimum.
And
For a free-fall of a body, there is an integral of mo-
tion (quantity which is conserved), 1/2v^2 − gh, where v is an
instantaneous speed, and h is the current height of the body.

(I don't have slightest idea what to do after )
any geometric approach??

View attachment 258772

What if the stone started at point $c$?

 

[IamVector]

What if the stone started at point $c$?

I reduced it by taking c = 0 what after??

 

[IamVector]

What if the stone started at point $c$?

yeah it can but it will be just reverse of starting from a i think.

 

[IamVector]

What if the stone started at point $c$?

also,
Each parabola has a focus, the properties of which
are most easily expressed in terms of geometrical optics: if
the parabola reflects light, all those rays which are parallel
to the symmetry axis are reflected to the focus
due to the Fermat’ principle, this means also
that for each point on the parabola the distance to the focus
plus the distance to the infinitely distant light source are equal.
can it be used somehow?

 

[PeroK]

also,
Each parabola has a focus, the properties of which
are most easily expressed in terms of geometrical optics: if
the parabola reflects light, all those rays which are parallel
to the symmetry axis are reflected to the focus
due to the Fermat’ principle, this means also
that for each point on the parabola the distance to the focus
plus the distance to the infinitely distant light source are equal.
can it be used somehow?

The problem shouldn't be too hard with $c$ equal $0$. Solve that first and then worry about $c \ne 0$.

 

[IamVector]

The problem shouldn't be too hard with $c$ equal $0$. Solve that first and then worry about $c \ne 0$.

should I take launch angle 45 and then try to solve v with it??

 

[PeroK]

should I take launch angle 45 and then try to solve v with it??

The launch angle is not necessarily 45 degrees. It will depend on $a$ and $b$.

 

[IamVector]

The launch angle is not necessarily 45 degrees. It will depend on $a$ and $b$.

wouldn't there be 2 variables to solve then v and angle say 'α' at the end I have to represent v in α or α in v.

 

[PeroK]

wouldn't there be 2 variables to solve then v and angle say 'α' at the end I have to represent v in α or α in v.

Yes, it isn't that easy actually. You might need a few tricks. I don't see an easy way.

 

[IamVector]

Yes, it isn't that easy actually. You might need a few tricks. I don't see an easy way.

yep I think so

 

[IamVector]

Yes, it isn't that easy actually. You might need a few tricks. I don't see an easy way.

well can we use parabola property in #5 somehow?

 

[PeroK]

well can we use parabola property in #5 somehow?

Maybe. I just crunched it with some calculus!

PS It all miraculously simplifies at the end. There must be a quick way.

 

[sysprog]

When considering projectiles and slope, one should not neglect cosines -- e.g. for a 45 degree angle, gravity operates on only .7071 of the total distance, so you have to reduce the moa of your barrel elevation accordingly when you're aiming your shot.

 

[IamVector]

Maybe. I just crunched it with some calculus!

PS It all miraculously simplifies at the end. There must be a quick way.

at the end it all reduce to equation with both v and angle α whenever I substitute for range = d.

 

[IamVector]

Maybe. I just crunched it with some calculus!

PS It all miraculously simplifies at the end. There must be a quick way.

how by calculus?

 

[PeroK]

at the end it all reduce to equation with both v and angle α whenever I substitute for range = d.

Here's a strategy:

1) Work out the equation of any projectile parabola that goes through the point $(d, a)$. Fired from point $(0,0)$.

2a) Express $v^2$ in terms of $\alpha$.

3a) Minimise $v^2$ with respect to $\alpha$. Trick: set $u = \tan \alpha$.

4a) Simplify expression for $v^2$.

Alternative:

2b) For a given $v^2$, maximise the height $y$ when $x = d$.

3b) Set this maximum height to $a$.

4b) Simplify.

This results in some simpler algebra, but might be trickier to see that this is the same problem.

 

[IamVector]

Here's a strategy:

1) Work out the equation of any projectile parabola that goes through the point $(d, a)$. Fired from point $(0,0)$.

2a) Express $v^2$ in terms of $\alpha$.

3a) Minimise $v^2$ with respect to $\alpha$. Trick: set $u = \tan \alpha$.

4a) Simplify expression for $v^2$.

Alternative:

2b) For a given $v^2$, maximise the height $y$ when $x = d$.

3b) Set this maximum height to $a$.

4b) Simplify.

This results in some simpler algebra, but might be trickier to see that this is the same problem.

I used another approach geometrical one

so we know sum of distance from each point from focus and height is same in parabola also assuming the focus to be a cannon by same velocity u we can reach any point in parabola if angle is changed so h+h = b+a-c
thus u = √2gh substituting h we get u=√g(b+a-c)

 

[IamVector]

Here's a strategy:

1) Work out the equation of any projectile parabola that goes through the point $(d, a)$. Fired from point $(0,0)$.

2a) Express $v^2$ in terms of $\alpha$.

3a) Minimise $v^2$ with respect to $\alpha$. Trick: set $u = \tan \alpha$.

4a) Simplify expression for $v^2$.

Alternative:

2b) For a given $v^2$, maximise the height $y$ when $x = d$.

3b) Set this maximum height to $a$.

4b) Simplify.

This results in some simpler algebra, but might be trickier to see that this is the same problem.

I will try with algebra now :)

 

[IamVector]

Here's a strategy:

1) Work out the equation of any projectile parabola that goes through the point $(d, a)$. Fired from point $(0,0)$.

2a) Express $v^2$ in terms of $\alpha$.

3a) Minimise $v^2$ with respect to $\alpha$. Trick: set $u = \tan \alpha$.

4a) Simplify expression for $v^2$.

Alternative:

2b) For a given $v^2$, maximise the height $y$ when $x = d$.

3b) Set this maximum height to $a$.

4b) Simplify.

This results in some simpler algebra, but might be trickier to see that this is the same problem.

is your answer √g(b+a+c) or √g(b+a-c)?? just want to confirm cause correct answer is √g(b+a+c)
and I am getting √g(b+a-c)

 

[PeroK]

is your answer √g(b+a+c) or √g(b+a-c)?? just want to confirm cause correct answer is √g(b+a+c)
and I am getting √g(b+a-c)

It's $a + b - c$ when you start from a height of $c$, and $a + b + c$ when you start from the ground.

 

[IamVector]

It's $a + b - c$ when you start from a height of $c$, and $a + b + c$ when you start from the ground.

oops totally forgot thanks :)

 

[PeroK]

I used another approach geometrical one
View attachment 258792
so we know sum of distance from each point from focus and height is same in parabola also assuming the focus to be a cannon by same velocity u we can reach any point in parabola if angle is changed so h+h = b+a-c
thus u = √2gh substituting h we get u=√g(b+a-c)

How do you know the focus is at the height $c$?

 

[IamVector]

How do you know the focus is at the height $c$?

I thought that velocity is constant so just by changing angle we can reach any point I took help of this diagram

 

[IamVector]

I thought that velocity is constant so just by changing angle we can reach any point I took help of this diagram
View attachment 258794

its from projectile motion and just by changing angle its touching every part of parabola .

 

[IamVector]

its from projectile motion and just by changing angle its touching every part of parabola .

except from 0-45