Free-Fall Physics Problem: Position, Height, Time, and Speed of a Falling Brick

In summary, the problem involves a crane lifting a load of bricks at a steady velocity of 5.0 m/s. One brick falls off from a height of 6.0 m above the ground. The task is to sketch the position of the brick y(t) versus time, determine the greatest height it reaches, calculate the time it takes to reach the ground, and find its speed just before hitting the ground. To solve this problem, we can use the formula d = di + Vi*t + ½*a*t², where d is the distance, di is the initial distance, Vi is the initial velocity, a is the acceleration, and t is the time. By plugging in the given values, we can find the
  • #1
pjr65stang
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Homework Statement


A load of bricks is lifted by a crane at a steady velocity of 5.0 m/s when one brick falls off 6.0 m above the ground. (a) Sketch the position of the brick y(t) versus time, from the moment it leaves the pallet until it hits the ground. (b) What is the greatest height the brick reaches above the ground? (c) How long does it take to reach the ground? (d) What is its speed just before it hits the ground?


Homework Equations





The Attempt at a Solution


I'm not sure where to even begin. The answers are in the back of the book, and the graph for part A is completely different than I thought. A time interval isn't given. I'm just really unsure of what to do here.
 
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  • #2


Welcome to PF!
We use formulas to describe the motion of a moving object.
When there is no acceleration, the formula is just d = v*t, where d is the distance, v the constant speed and t the time.
In this problem, there is the acceleration of gravity so the distance formula is more complicated: d = di + Vi*t + ½*a*t²
If you put in the initial distance di = 6, the initial speed for Vi and the acceleration of gravity -9.81 for "a", you can use the formula to find the height d at any given time. Do it for every half second from 0 to 3 or 4 seconds and you'll have enough points to graph the height vs time and answer the other questions.
 

1. What is free-fall in physics?

Free-fall in physics refers to the motion of an object under the sole influence of gravity, with no other forces acting upon it. This means that the object is accelerating towards the ground at a constant rate of 9.8 meters per second squared.

2. How is free-fall different from normal falling?

In free-fall, the only force acting on the object is gravity, whereas in normal falling, there may be other forces at play, such as air resistance. This means that objects in free-fall will accelerate at a constant rate, while objects in normal falling may experience fluctuations in their acceleration due to other forces.

3. What is the formula for calculating free-fall acceleration?

The formula for calculating free-fall acceleration is a = g, where a is the acceleration and g is the acceleration due to gravity, which is approximately 9.8 meters per second squared on Earth.

4. How is the distance traveled in free-fall calculated?

The distance traveled in free-fall is calculated using the formula d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity, and t is the time in seconds.

5. Can an object reach terminal velocity during free-fall?

Yes, an object can reach terminal velocity during free-fall. Terminal velocity is the maximum speed an object can reach when falling, when the force of gravity is balanced by the force of air resistance. In a vacuum, where there is no air resistance, an object will continue to accelerate until it reaches the ground.

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