# Kinematics of a particle

1. Oct 29, 2016

### Kajan thana

1. The problem statement, all variables and given/known data
A ball is thrown horizontally wit speed 20m/s, from the top of the building which is 30m high.
Find the time the ball takes to reach the ground?

2. Relevant equations
S= ut+0.5at^2
3. The attempt at a solution
S=-30
u=0
A=9.8
T=?

When I pluck the values I will get t^s=-30/4.9 but when we square root we will get complex number, but if I take the displacement as +30 then it will give me a positive value, so I am confused on why the s is not negative.

If I say the displacement where the ball is on top of the building, then the displacement is 0 but if it goes down then it should be negative value.

2. Oct 29, 2016

### TomHart

I think if I was working this problem, to make it easiest, I would define the down direction as positive (that way acceleration and displacement will both be positive) and the top of the building as x=0. Or you could define down as negative and acceleration and displacement would be negative. But it's for sure that if you throw a ball off of a building, the time required for the ball to hit the ground will be a positive value.

3. Oct 29, 2016