(adsbygoogle = window.adsbygoogle || []).push({}); 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

Quadratic equation

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

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# Kinematics of a particle motion equations problem

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