# Kinematics of a particle motion equations problem

1. Dec 31, 2011

### Rob K

1. The problem statement, all variables and given/known data
A baseball is thrown downward from a 12.5m tower with and initial speed of 4.5m/s. Determine the speed at which it hits the ground and the time of travel.
s = 12.5m
u = 4.5m/s
a = g = 10m/s^2

2. Relevant equations

Motion equations used
s = ut + 0.5at^2
v^2 = u^2 +2as

x = (-b +/- sqrt( b^2 - 4ac))/2a

3. The attempt at a solution

This is what I started with, and rearranged to put into the quadratic formular.
12.5 = 4.5t + 0.5(10)t^2
therefore:
5t^2 + 4.5t - 12.5 = 0

Where:
a = 5
b = 4.5
c = 12.5

The answer I get is 1.194 seconds
The answer in the book says exactly half that at 0.597 seconds.

Then for part two:
v^2 = u^2 + 2as
seemed the right one.

and I consistently get 16.439m/s
yet the answer in the book says 23.9m/s

I would be grateful for any input on this.

Kind Regards

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

2. Relevant equations

3. The attempt at a solution

2. Dec 31, 2011

### nvn

Rob K: Both of your answers currently appear to be correct, and the book appears to be wrong.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 12.5 m, not 12.5m. See the international standard for writing units (ISO 31-0).

3. Dec 31, 2011

### Rob K

Thank you very much indeed for verifying this for me, I am suspecting there are a few errors, though as it is the 11th edition, I thought they might have been ironed out by now.

Happy New Year

Rob K