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Using Kinematics to find the Distance of an object dropped from rest

  • Thread starter garcia1
  • Start date
1. The problem statement, all variables and given/known data

A stone is thrown straight up from the ground
with an initial speed of 44.1 m/s . At the same
instant, a stone is dropped from a height of
h meters above ground level. The two stones
strike the ground simultaneously.
Find the height h. The acceleration of
gravity is 9.8 m/s2 .
Answer in units of m.


2. Relevant equations

Kinematics equations, such as V^2 = Vo^2 + 2ax, V = Vo +at, x = Vot +1/2at^2, and x = 1/2(Vo +V)t


3. The attempt at a solution

I first thought that by plugging the known variables, with V as 0 m/s, into V^2 = Vo^2 +2ax, solving for x. I got x=99.225. This was wrong. I then re-read the statement, thinking that I needed to double my answer, since the object thrown up also had a downward portion before it hit the ground with the falling object. This was x = 198.45m. This was also wrong.
 
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First, you need to find the time it takes for the rock thrown upwards to hit the floor again. Using that time, you know that the second rock covered a distance of h in the same amount of time under uniform acceleration.
 

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