Kinematics of a particle motion equations problem

[Rob K]

Homework Statement

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

Homework Equations

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

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

The Attempt at a Solution

This is what I started with, and rearranged to put into the quadratic formula.
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

 

[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[/color] (ISO 31-0[/color]).

 

[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