Kermit's Problem (1 Viewer)

[OLDMAN]

A frictionless* frog jumps from the ground with speed V at an unknown angle to the horizontal. It swallows a fly at a height h. Show that the frog should position itself within a radius of
V/g SQRT(V^2-2gh) of the point below gulp point.

This problem should come with a warning and a promise.
Warning :Quite difficult. It is a dangerous swamp out there, and the problem poses a few traps.
Promise : an earnest attempt yields deeper insights on projectiles and polynomials.

* Yeah, super slimy.

 

[ND]

Originally posted by OLDMAN
Show that the frog should position itself within a radius of
V/g SQRT(V^2-2gh) of the point below gulp point.

I don't understand what it's asking. Is this where it starts from? Is that where it is when it eats the fly?

edit: nevermind, i think i get it now.

 

[ND]

Using the physics formulas:

d = t*v_x ...[1] (where v_x is the initial horizontal velocity)
h = v_y*t - (gt^2)/2 ...[2]

now you can find v using pythag:
v^2 = (v_x)^2 + (v_y)^2

rearranging 1 and two and summing together gives:
d^2/t^2 + (4h^2 + 4hgt^2 + g^2*t^4)/4t^2 = v^2
g^2*t^4 + 4(gh - v^2)t^2 + 4(h^2 + d^2) = 0

now for a real root:
16(gh - v^2)^2 - 16g^2(h^2 + d^2) >= 0
d <= (v/g)*sqrt(v^2 - 2gh)

 

[Exeter]

if the frong was frictionless it wouldnt be able to jump forward, only up.

 

[ND]

 

[OLDMAN]

Very good, ND.

Just a note on the Physics of Polynomials!? (as promised).

0<=d <= (v/g)*sqrt(v^2 - 2gh) give the boundaries of d. Could it then be deduced that maximum d=(v/g)*sqrt(v^2 - 2gh) is actually possible?
The answer is yes. Since 1/2 v^2>gh (conservation of energy eqn., otherwise Kermit won't reach no matter what angle)),
then 4(gh - v^2)<0
or -4(gh - v^2)>0 in the polynomial
g^2*t^4 + 4(gh - v^2)t^2 + 4(h^2 + d^2) = 0

This implies that there is a quadratic solution,
t^2=-4(gh - v^2)+-etc. (quadratic formula), and hence a solution for t>0.
What is interesting is Kermit catches the fly on the way down, and not when v_y=0 ie. a projectile path where h is the maximum height. btw, that distance is only d=(V_y/g)*sqrt(v^2 - 2gh) which is smaller. So Kermit peaks before the gulp, sacrficing a little bit of horizontal velocity for more flight time.

Very impressed with the quality of the students active in this forum.

 

[OLDMAN]

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Quote Exeter:

if the frog was frictionless it wouldnt be able to jump forward, only up.
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Forgot to mention "frictionless except soles of feet". While in the topic of Frog Physics, anyone seen Peter Brock's frog tire ad. A clever frog indeed, lying flat on the bonnet to minimize air resistance, raising its head a bit to create an aerodynamic negative lift to increase the normal force thereby increasing friction. Sadly, in the end Newton's First Law caught up with slimy and it became a test of what had more friction - frog feet on shiny metal or rubber on tarmac.