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

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SUMMARY

The problem involves two stones: one thrown upwards with an initial speed of 44.1 m/s and another dropped from a height h, both striking the ground simultaneously. The acceleration due to gravity is 9.8 m/s². To find the height h, the time taken for the first stone to reach the ground must be calculated using kinematics equations. The correct approach involves determining the time for the upward stone's motion and then applying that time to calculate the distance h for the dropped stone.

PREREQUISITES
  • Understanding of kinematics equations, specifically V^2 = Vo^2 + 2ax and x = 1/2(Vo + V)t
  • Knowledge of the concepts of free fall and projectile motion
  • Ability to solve quadratic equations
  • Familiarity with the acceleration due to gravity (9.8 m/s²)
NEXT STEPS
  • Calculate the time of flight for an object thrown upwards using kinematics
  • Explore the relationship between time of flight and distance in free fall scenarios
  • Study the derivation and application of kinematic equations in various motion problems
  • Investigate the effects of initial velocity on the trajectory of projectiles
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for examples of real-world applications of these concepts.

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Homework Statement



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.


Homework Equations



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


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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