Kinematics - Two particles release from same height ...

Good afternoon (from Europe) !

The given solution of mentioned problem is wrong.
There must be some initial velocity of particle "A" moving downwards along slope.

Particle "\"B\"": t_B=√(2h/g)
Particle "\"A\" ": L=v_(A,i) t+a t^2/2=v_(A,i) √(2h/g)+gsinα (√(2h/g))^2/2=v_(A,i) √(2h/g)+h^2/L

v_(A,i)=L(1-h^2/L^2 ) √(g/2h)=20*(1-〖12〗^2/〖20〗^2 ) √(9.81/(2*12))=8.18 m/s

The result meets our expectation.

Best regards !

congestus

Solid mensuration: a modest contribution to a solution 016

Example 016: Radius of sphere circumscribing a regular triangular pyramid

I would like to add a modest contribution to a solution 016.
Radius of inscribbed sphere can be obtained directly (without slope angle determination):

√((AE)^2+(OE)^2 ) = ED-OE → OE = r = ((ED)^2-(AE)^2)/(2*(ED) ) and so on

Best regards !

Matija Oblak