Find the speed v0 and height

Question

The particle thrown under the angle pi/6 rad. with the horizontal direction achieves the height h for t1=10S when going up and t2=50S when going down. Find initial speed v0 and height h.

As h1=h2
v0y*t1-gt1^2/2=gt2^2/2
v*0.5*10-10*100/2=10*2500/2
v0=2600. 
hmax:
v^2-vy0^2=2gh
vy0=v0*sin30=2600*0.5=1300
h=0-1300^2=2*(-10)*h
h=84500m.
Am i right?

Comments

This is not graduate-level, voting to close.