I hope someone can help me, I am completely confused at the moment. I'm working on Projectiles with the rules of:

Vertical:

Ay = -9.8m/s^

Vfy = Viy + at

Change Y = Viy t + 1/2 ayt^

Horizontal:

Ax = 0m/s^

Vx = Change of X / t

Question: A grenade is thrown off from the top of a cliff at an angle of 25 degrees with an initial speed of 15 m/s. After 3 seconds in flight, how far will it have fallen below the ledge.

I have to show my work so this is what I have so far

Viy = 15 m/s(sin25) = 6.34m/s

Vi = 6.34 m/s
a = -9.8 m/s^
t = 3 sec

So i went with Vf = Vi + at

6.34m/s + (-9.8m/s^) 3
= 23.06

the book says the answer is 25.08 so I'm confused on where I went wrong... If you can help I'd appreciate it.

Sorry about the double post.. got an error while posting and for some reason it put it up twice..

I get the same vertical speed as you.

Not even the speed's module (Ay^2+Az^2)^0.5 does match 25.08.

Thank you, I'm thinking the book is wrong... which means I just spent a lot of time trying to figure it out, when I already did! I appreciate it!

What confuses me is that the question asks for a distance and you give a velocity.

The equation you want is
s= ut +(1/2)at^2
If we measure distance down then your initial velocity (up) is negative
so..
-6.34*3 + (1/2)x9.8x9
?