How Fast Must a Book Be Tossed to Clear a Windowsill?

AI Thread Summary
To clear a windowsill 4.2 meters high, a book must be tossed with an initial velocity of approximately 7.27 m/s from a height of 1.5 meters. After being thrown, the book will hit the floor 0.87 meters below the windowsill in about 1.16 seconds. Two potential times were calculated, but the later time corresponds to when the book impacts the ground. The discussion emphasizes the importance of defining a consistent origin point and direction for displacement in the calculations. Overall, the physics principles applied clarify the motion of the book in relation to the defined heights.
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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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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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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.
 
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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