Height of tower given objects dropped from top at same time?

[Ab17]

Homework Statement

A stone is thrown vertically upward from the top of the tower with a velocity of 15m/s. Two seconds later a second stone is dropped from the top of the tower.if both the stone strike the ground simultaneously find the height of the towerV1i=15m/s
V2i=0m/s
t=2s
h=?

Homework Equations

Xf=Xi+0.5at^2
Vf=vi+at

The Attempt at a Solution

I used two methods and got different answers which is confusing me

METHOD 1:
Xf1 = 15t-4.9t^2
Xf2 = -4.9(t-2)^2
Xf2 = -4.9t^2 +16.9t -16.9

Xf2=Xf1
15t -4.9t^2 = -4.9t^2 +16.9t -16.9
-1.9t = -16.9
t= 8.89s
h=253.90 m

METHOD 2:
Xf1= 15(2) -4.9(2)^2
Xf1 = 10.4m
Vf= Vi + at
Vf= 15 -9.8(2)
Vf = -4.6m/s
Xf1= 10.4 - 4.6t -4.9t^2
Xf2 = -4.9t^2

Xf1=Xf2
10.4 -4.6t -4.6t^2 = -4.9t^2
10.4=4.6t
t = 2.26s
h= 25.03m

Which solution is right?Why am I getting different answers? Is it because the 15m/s velocity is upwards and will reach maximum height? I am really confused

 

[CWatters]

I haven't check it all but...

Xf2 = -4.9(t-2)^2
Xf2 = -4.9t^2 +16.9t -16.9

4.9 * 4 = 19.6 not 16.9

 

[Ab17]

Thank you

 

[Ab17]

Sometimes you get the hard things right but make mistakes on the easy things