Laws of Motion- dropping book off building with a twist

[a.k]

Homework Statement

A 2.0kg physics book is dropped from the roof of a skyscraper 240.1 m high. While the book is falling to the ground, a horizontal wind exerts a constant force of 11 N on it. Ignore air resistance.

a. How far from the building is the book when it hits the ground?
b. How long does it take the book to hit the ground?
c. What is the speed of the book when it hits the ground?

Homework Equations

a_y=-g=-9.8m/s^2
a_x=force of the wind/m

The Attempt at a Solution

Using the second formula,

11N/2kg=5.5 m/s^2

I would like some advice as to what to do now.

 

[a.k]

Ok, I figured out some more info. on this problem.

For part b,

t=sqrt(2y/ay)

2*-240.1

sqrt(49)

t=7 secs

 

[Doc Al]

Ok, I figured out some more info. on this problem.

For part b,

t=sqrt(2y/ay)

2*-240.1

sqrt(49)

t=7 secs

Good. Now use the horizontal acceleration you calculated to solve part a.

 

[a.k]

So part a,

x=1/2axt^2

1/2(5.5)(7)^2
269.5/2
x=134.75 m

Im not sure if using "x" is correct. It is suppose to show horizontal accel, so ax?

 

[Doc Al]

So part a,

x=1/2axt^2

1/2(5.5)(7)^2
269.5/2
x=134.75 m

Good.

Im not sure if using "x" is correct. It is suppose to show horizontal accel, so ax?

Calling the horizontal component of acceleration a_x is fine.

 

[a.k]

Ok I am working on part c and ax is actually 5.5 m/s^2. I am confusing myself.

 

[Doc Al]

Ok I am working on part c and ax is actually 5.5 m/s^2. I am confusing myself.

Find the vertical and horizontal components of the velocity when it hits the ground.

 

[a.k]

Vy=ayt
Vx=axt

Vy=-68.6 m/s
Vx=38.5 m/s

V=sqrt(38.5^2+-68.6^2)
sqrt(6188.21)
V=78.67 m/s

 

[Doc Al]

Good job!