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# Projectile question?? watch

1. There is a hemispherical hill with radius R. A Football player kicks a ball placing on the top of the hill horizontally. Find the minimum velocity v that the ball will never touch the hill again after being kicked. And determine a distance measuring from the centre of the hemispherical hill for that minimum velocity.

Neglect air resistance.
Only Projectile Motion
2. (Original post by helbrida)
There is a hemispherical hill with radius R. A Football player kicks a ball placing on the top of the hill horizontally. Find the minimum velocity v that the ball will never touch the hill again after being kicked. And determine a distance measuring from the centre of the hemispherical hill for that minimum velocity.
For the ball not to touch the hill again, it will need to travel a distance equal to or greater than R, in its horizontal motion. The velocity will only have one component as it is kicked in the horizontal direction. So we are finding a horizontal velocity.

v = Minimum Distance Required/Time = R/T

The ball's starting height is R, so, in modelling its vertical motion:

s = R, a = 9.8, u = 0

s = ut + 1/2at^2
R = 4.9t^2

t = sq.root (R/4.9)

Hence: v = R/T = R / [sq.rt (R/4.9)]
= (R * sq.root4.9)/(sq.root R) = sq.root (4.9R)

---> v = sq.root (4.9R)

Measuring Distance = Minimum Velocity * Time = sq.root (4.9R) * sq.root (R/4.9) = sq.root (4.9R^2 / 4.9) = R

---> Measuring Distance (From Centre Of Hill) = R

NB: I assumed that from the wording of the question, touching the outside edge of the hill is effectively permitted as *a* minimum velocity is required, and hence the case where the ball touches the outside edge of the hill would need to be considered.

Otherwise, we would have the two conditions:

v ≥ sq.root (4.9R)
Measuring Distance (From Centre Of Hill) ≥ R
3. Does hemispherical mean that the top will always be at the mid-point of the hilll. Apologies, I am unfamiliar with the term.
4. (Original post by RobbieC)
Does hemispherical mean that the top will always be at the mid-point of the hilll. Apologies, I am unfamiliar with the term.
Yes, I think so.
5. (Original post by RobbieC)
Does hemispherical mean that the top will always be at the mid-point of the hilll. Apologies, I am unfamiliar with the term.
Yeah it does. A hemisphere is just half a sphere.
6. (Original post by Nylex)
Yeah it does. A hemisphere is just half a sphere.
Ralfskini said that my solutions were correct.
7. Hi lads. I think there's a problem here.
If the ball only travels a (horizontal) distance R, in the time it takes to drop a vertical distance R, then the ball will INTERSECT the hill! See Fig1.

The eqns of motion are.

Sx = vt
Sy = R - ½gt²

Sy = R - (g/2v²)Sx² - the projectile path

if v = √(gR/2), then

Sy = R - (1/R)Sx²
============

I've plotted Sy = R - (1/R)Sx² and x² + y² = R² (the hill) in Fig1. As you can see, they intersect. (To plot the graphs, I set R=20)
The initial velocity v will have to be greater than √(gR/2) to ensure the curves Sy = R - (g/2v²)Sx² (the projectile path) and x² + y² = R² (the Hill) do not intersect. i.e. that the ball clears the hill.
This condition is satisfied by making the radius of curvature of the graph of Sy = R - (g/2v²)Sx² greater than the radius of curvature of x² + y² = R² - which of course is R.

The projectile path can be written in parametric form as,

x = 2at , y = R - at²

The curvature is given by,

ρ = (x'y'' - y'x'')/(x'² + y'²)^(3/2)

where x' = dx/dt, y' = dy/dt, etc.

x' = 2a, x'' = 0
y' = -2at, y'' = -2a

ρ = (2a.(-2a) - 0)/(4a² + 4a²t)^(3/2)
ρ = -4a²/[8a³(1+t²)^(3/2)]

at t=0, (which is the kick-off point) ρ = -4a²/8a³ = -½a
Radius of curvature = |1/ρ| = 2a

So,
R = 2a
a = R/2
=====

Then, (using the parametric equations)

x = Rt, y = R - (R/2)t²
y = R - (1/2R)x²

But we have Sy = R - (g/2v²)Sx²

Therefore,

2R = 2v²/g
v² = gR
v = √(gR) - See Fig2
=======

The projectile path is,

y = R - (1/2R)x²

At y = 0,

x² = 2R²
x = √(2)R
========
Attached Images

8. (Original post by Fermat)
Hi lads. I think there's a problem here.
If the ball only travels a (horizontal) distance R, in the time it takes to drop a vertical distance R, then the ball will INTERSECT the hill! See Fig1.

The eqns of motion are.

Sx = vt
Sy = R - ½gt²

Sy = R - (g/2v²)Sx² - the projectile path

if v = √(gR/2), then

Sy = R - (1/R)Sx²
============

I've plotted Sy = R - (1/R)Sx² and x² + y² = R² (the hill) in Fig1. As you can see, they intersect. (To plot the graphs, I set R=20)
The initial velocity v will have to be greater than √(gR/2) to ensure the curves Sy = R - (g/2v²)Sx² (the projectile path) and x² + y² = R² (the Hill) do not intersect. i.e. that the ball clears the hill.
This condition is satisfied by making the radius of curvature of the graph of Sy = R - (g/2v²)Sx² greater than the radius of curvature of x² + y² = R² - which of course is R.

The projectile path can be written in parametric form as,

x = 2at , y = R - at²

The curvature is given by,

ρ = (x'y'' - y'x'')/(x'² + y'²)^(3/2)

where x' = dx/dt, y' = dy/dt, etc.

x' = 2a, x'' = 0
y' = -2at, y'' = -2a

ρ = (2a.(-2a) - 0)/(4a² + 4a²t)^(3/2)
ρ = -4a²/[8a³(1+t²)^(3/2)]

at t=0, (which is the kick-off point) ρ = -4a²/8a³ = -½a
Radius of curvature = |1/ρ| = 2a

So,
R = 2a
a = R/2
=====

Then, (using the parametric equations)

x = Rt, y = R - (R/2)t²
y = R - (1/2R)x²

But we have Sy = R - (g/2v²)Sx²

Therefore,

2R = 2v²/g
v² = gR
v = √(gR) - See Fig2
=======

The projectile path is,

y = R - (1/2R)x²

At y = 0,

x² = 2R²
x = √(2)R
========

Good point. I didn't actually think about the path of the ball (I didn't really think much about the question) but just checked that it would pass through (R,0) (clearly it will hit the hill before this).

I may as well have another go at this using a slightly different method to that of Fermat:

x²+y²=R²
y=R-gx²/2v²

We need to find the minimum value of v such that these two curves do not intersect.

√(R²-x²)=R-gx²/2v²
R²-x²=(R-gx²/2v²)²=R²-2Rgx²/2v²+g²x^4/4v^4
g²x^4/4v^4+x²(1-2Rg/2v²)=0
g²x²/4v^4+(1-2Rg/2v²)=0

no real roots if -g²(1-2Rg/2v²)/v^4>0
Rg^3/v^6-g²/v^4>0
v>√(Rg)

..which is probably wrong because I'm too tired to do this.
9. (Original post by Ralfskini)
...

x²+y²=R²
y=R-gx²/2v²

We need to find the minimum value of v such that these two curves do not intersect.

√(R²-x²)=R-gx²/2v²
R²-x²=(R-gx²/2v²)²=R²-2Rgx²/2v²+g²x^4/4v^4
g²x^4/4v^4+x²(1-2Rg/2v²)=0
g²x²/4v^4+(1-2Rg/2v²)=0

no real roots if -g²(1-2Rg/2v²)/v^4>0
Rg^3/v^6-g²/v^4>0
v>√(Rg)

..which is probably wrong because I'm too tired to do this.
... which is right cos' I just checked it!

Neat solution. Much quicker than mine.
Just to finish off, ...

using y=R-gx²/2v² to find where the ball hits the ground,

R = gx²/2v² (at y=0)
x² = (2R/g)v²
v² > Rg
x² > (2R/g)(Rg)
x² > 2R²
x > √(2)R
10. Thanks for answering my stupidity of question.
With great respect.

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