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
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

Answer 1

nope, v=d/t is average velocity. Was there something you did not understand on how I told you to do it?

Answer 2

Yes, if you go back to the post I don't understand the sqart or that whole equation at all, grade 11 physics! Sorry for replying late btw