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

DYNAMICS

A balloon rises from the ground with a constant acceleration of 4ft per sec^2. Five seconds later, a stone is thrown vertically up from the launching site. What must be the minimum initial velocity of the stone for it to just touch the balloon? Note that the balloon and stone have the same velocity at contact.

truss: simply supported right-angle triangle truss

A simple supported right-angle triangle truss of 1/2( 5 x 12 ) meter square, is subjected with 50KN and 25KN at the mid-span and top end of the truss respectively. The truss is having two internal members one acting horizontal at the mid-span and the other acting diagonal from support A to joint at a point on the slanted side of right angle triangle.
Taking into account that the pined support is on the right and the roller on the left.
You are required to check the determinant and compute the reactions of this truss.