Question

A soccer ball is kicked with a velocity of 12m/s, HORIZONTALLY off a building which is 8.5m high. What is the velocity of the ball after 1.25s?

This is what I found so far:
Vi = 12m/s
Dx = 15.8049336m
Dy = 8.5m
Time it took for ball to hit floor 1.31707778s
I used D=Vi(T) + 1/2a (T)^2 to find total time
I used dX = Vi(T) to find the range
Thank you!

Answer 1

vertical velociy= 16t
so figure that at t=1.25sec

then
velocity=sqart(12^2 + vertVel^2)

and you have it

Now if you want to find the direction of the velocity, you have to figure that, I assume you just want speed.

Answer 2

I am sorry I do not know what sqart is, and also sorry for not replying sooner bobpursley, but I tried to solve it myself and this is what I did, perhaps I used sqart?

Because V = D/T I decided to find Dx and Dy
Found Dy using:
D= Vi(1.25) + 1/2(9.8)1.25^2
Which gave me 7.65625m vertical distance at 1.25s

Found Dx using:
Dx = Vi(1.25) +1/2(0)1.31707778
Which gave me 15m horizontal distance at 1.25s

Because I have vertical and horizontal distance I made a triangle which would give me the velocity? Or distance?
This is what I did:
7.65625^2 + 15^2 = 16.84096684^2

Using 16.84096684m I did V = D/T and got this:
16.84096684m/1.25s = 13.47277347m/s

I did this before you gave me my answer but appreciate it anyways, can you tell me if I did it correctly? Im in grade 11 physics and have never heard of the term sqart so im not sure exactly if I did what you did.