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## 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 tower

V1i=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? Im really confused