# Laws of Motion- dropping book off building with a twist

1. Mar 20, 2014

### a.k

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
ay=-g=-9.8m/s^2
ax=force of the wind/m

3. 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.

2. Mar 20, 2014

### a.k

For part b,

t=sqrt(2y/ay)

2*-240.1

sqrt(49)

t=7 secs

3. Mar 20, 2014

### Staff: Mentor

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

4. Mar 20, 2014

### 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?

5. Mar 20, 2014

### Staff: Mentor

Good.

Calling the horizontal component of acceleration ax is fine.

6. Mar 20, 2014

### a.k

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

7. Mar 20, 2014

### Staff: Mentor

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

8. Mar 20, 2014

### 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

9. Mar 20, 2014

Good job!