# Given height acceleration

1. Feb 26, 2012

### jkmath16

1. The problem statement, all variables and given/known data

ball thrown horizontally from roof of building 55.0m tall and lands 29.7m from the base. find balls initial speed

2. Relevant equations
(√acceleration/[√2*55] ) * 29.7=8.86

3. The attempt at a solution
(sqroot acceleration/[2*55] ) * 29.7=8.86
the answer is correct..I am confused how it is that i can get to this formula.

I am confused to what "29.7m from the base" means
thanks all
*this is my first time posting, I hope i am doing it correct

2. Feb 26, 2012

### PeterO

This is an example of projectile motion.
The vertical part of the motion is exactly the same as vertical motion under gravity - especially with respect to time.
The horizontal component is just motion at constant velocity.

If you drop a ball 55m, it takes a little over 3 seconds to reach the ground. This ball will also take that same time to reach the ground.
In that time it had been moving away from the wall at constant speed.

Lets pretend it took exactly 3 seconds to land.
If it was initially traveling at 8 m/s, it will land 24m from the wall after 3 seconds.

Your ball tookmore than 3 seconds to get there, and landed 29.7 metres out, but you should be able to do the arithmetic.