Book toss Kinematics motion

In summary: The important thing for the displacement is that you define an origin point (in this case, it's the floor directly below your hand) and a positive direction (in this case "up").So the initial displacement of the book is +1.5m. The final displacement is +3.33m (because it ends up 3.33m above the level of the dorm room ground).
  • #1
negation
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Homework Statement




A book is toss into your dorm room, just clearing a windowsill 4.2m above ground.
a)If the book leaves your hand 1.5m above ground, how fast must it be going to clear the sill?
b)How long after it leaves your hand will it hit the floor, 0.87m below the windowsill?



Homework Equations



none.

The Attempt at a Solution



a)vf^2 - vi^2 = 2g(yf - yi)

-vi^2 = 2(-9.8ms^-2) (4.2m-1.5m)
vi = 7.274613392ms^-1


b)vi = 7.274613392ms^-1
yf - yi = 4.2m - 1.5m (are the directions correct?)
g = -9.81ms^-2

yf - yi = vit + 0.5gt^2

3.33m - 1.5m = 7.274613392ms^-1 t - 4.905 ms^-2 t^2

solving via quadratic equation:

t = 1.162s && t = 0.32

Which t do I take? Conceptual explanation please?
 
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  • #2
negation said:

Homework Statement




A book is toss into your dorm room, just clearing a windowsill 4.2m above ground.
a)If the book leaves your hand 1.5m above ground, how fast must it be going to clear the sill?
b)How long after it leaves your hand will it hit the floor, 0.87m below the windowsill?



Homework Equations



none.

The Attempt at a Solution



a)vf^2 - vi^2 = 2g(yf - yi)

-vi^2 = 2(-9.8ms^-2) (4.2m-1.5m)
vi = 7.274613392ms^-1


b)vi = 7.274613392ms^-1
yf - yi = 4.2m - 1.5m (are the directions correct?)
g = -9.81ms^-2

yf - yi = vit + 0.5gt^2

3.33m - 1.5m = 7.274613392ms^-1 t - 4.905 ms^-2 t^2

solving via quadratic equation:

t = 1.162s && t = 0.32

Which t do I take? Conceptual explanation please?

The later one (1.162s). The earlier t signifies the time the book passes the spot 3.33m above the floor on its upward trajectory (on the way to the windowsill). The later t is the time the book hits the floor outside.
 
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  • #3
Curious3141 said:
The later one (1.162s). The earlier t signifies the time the book passes the spot 3.33m above the floor on its upward trajectory (on the way to the windowsill). The later t is the time the book hits the floor outside.

And I presume the directions of yf and yi in part b is both correct?

I would have made the assumption that at yi = 1.5m and going up, the displacement is positive and at yf = 3.33m, the displacement would be negative.
 
  • #4
negation said:
And I presume the directions of yf and yi in part b is both correct?

I would have made the assumption that at yi = 1.5m and going up, the displacement is positive and at yf = 3.33m, the displacement would be negative.

The important thing for the displacement is that you define an origin point (in this case, it's the floor directly below your hand) and a positive direction (in this case "up").

So the initial displacement of the book is +1.5m. The final displacement is +3.33m (because it ends up 3.33m above the level of the dorm room ground).

The problem is also completely workable if you take the origin as your hand. In which case the original displacement is 0, and the final displacement is +1.83m. The only thing that matters is consistency.
 
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  • #5


I would first clarify the given scenario and make sure all units are consistent. From the given information, we know that the book is thrown from a height of 1.5m and must clear a windowsill that is 4.2m above the ground. We are also given the acceleration due to gravity, which is -9.8 m/s^2.

To solve part a), we can use the kinematic equation vf^2 - vi^2 = 2a(yf-yi), where vf is the final velocity, vi is the initial velocity, a is the acceleration, and yf-yi is the change in position. We know that the final position (yf) is 4.2m and the initial position (yi) is 1.5m, so we can plug these values in and solve for vi.

For part b), we can use the kinematic equation yf-yi = vit + 0.5at^2, where t is the time it takes for the book to hit the ground. We know that the final position (yf) is 0.87m and the initial position (yi) is 1.5m, so we can plug these values in along with the calculated vi from part a) and solve for t.

As for which t to take, we must consider the physical meaning of the solutions. The first solution, t = 1.162s, would mean that the book takes 1.162 seconds to hit the ground after it is thrown. However, this solution does not make sense because it takes less time for the book to hit the ground than it does to clear the windowsill. Therefore, we would take the second solution, t = 0.32s, which means that the book takes 0.32 seconds to hit the ground after it is thrown. This makes sense as it takes longer for the book to clear the windowsill than it does to hit the ground.
 

1. What is "Book toss Kinematics motion"?

"Book toss Kinematics motion" refers to the study of the motion of a book as it is tossed through the air. This involves analyzing the book's position, velocity, and acceleration over time.

2. What are the key principles of kinematics motion?

The key principles of kinematics motion include displacement, velocity, acceleration, and time. These concepts are used to describe the motion of an object and how it changes over time.

3. How is the trajectory of a book toss calculated?

The trajectory of a book toss can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and time elapsed. These equations can be solved to determine the horizontal and vertical components of the book's motion.

4. What factors affect the motion of a book when tossed?

The motion of a book when tossed can be affected by several factors, including the initial velocity, air resistance, and the force of gravity. The shape and weight of the book can also impact its motion.

5. How is kinematics motion applied in real life?

Kinematics motion is applied in many real-life situations, such as calculating the trajectory of a ball in sports, predicting the motion of a projectile, or analyzing the movement of a car. It is also used in engineering and robotics to design and control machines and devices.

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