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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.
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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