Kermit's second problem. (1 Viewer)

[Affinity]

another Kermit's problem

ABC is a triangular obstacle, with AB horizontal.
This time the slimy frog jumps from point A, just grazes C and lands at B.
Show that it jumped at an angle of:

tan^-1[tan(A) + tan(B)]

to the horizontal

 

[OLDMAN]

Nice one Affinity.
Kermit reckons easier to prove general case -took off his Physics hat and put on the mathematics one instead.

 

[Affinity]

My maths teachers say:

"physics is an application of mathematics"
and
"Mathematics is a tool for physics"

one was a science teacher and another was a maths teacher

 

[OLDMAN]

Following proves the general case which should be sufficient.

A general parabola (x-h)^2=4a(y-k) could be written in parametric form as (h+2at , k+at^2).
If a< 0 and (h,k) is in the first quadrant then the parabola could represent a projectile path.
Let the end points of a horizontal chord (AB) be
(h+2aq,k+aq^2) and (h-2aq,k+aq^2) respectively (a <0). Let C be a point (h+2ap,k+ap^2) on the parabola in the first quadrant.

Then tan A=(p+q)/2 and tan B=|(p-q)/2| or tan B=(q-p)/2 since q>p, points A,B further from vertex than C, thus q>p.

Adding the results gives tan @=q which gives the intial projection angle.

Would like to see someone bash out the specific case using the projectile formulas.