Question

A student drops his textbook out of a window in a tall building. If it took 1.51 seconds for the book to hit the ground, how high above the ground is the window? What is the book' maximum speed during flight? At what time does this occur?

Answer 1

The window is at a height of 11.2 meters above the ground.

Explanation:

Ok, so let's see if I can do a better job explaining this.

Let's say that the window is alocated at a hight `h` above ground level. Since the student didn't throw the book from the window, which would imply that the book had an initial velocity, the book will free fall towards the ground.

During its flight, it will be accelerated by the gravitational acceleration, `g`. You know that it takes 1.51 seconds for the book to hit the ground.

This means that the distance `h` must have been

`h = underbrace(v_0)_(color(blue)(=0)) * t + 1/2 * g * t^2`

`h = 1/2 * 9.8"m"/color(red)(cancel(color(black)("s"^2))) * 1.51^2 color(red)(cancel(color(black)("s"^2))) = color(green)("11.2 m")`

Since the book keeps accelerating as it falls, it makes sense that it will have a maximum speed right before hitting the ground.

If `v` is the speed with which it hits the ground, and `h` is the distance it covered, then you can say that

`v^2 = underbrace(v_0^2)_(color(blue)(=0)) + 2 * g * h`

`v^2 = 2 * g * h implies v = sqrt(2 * g * h)`

`v = sqrt(2 * 9.8"m"/"s"^2 * "11.2 m") = color(green)("14.8 m/s")`

Alternatively, you can use the equation

`v = underbrace(v_0)_(color(blue)(=0)) + g * t`

which will produce the same value for `v`

`v = 9.8"m"/"s"^color(red)(cancel(color(black)(2))) * 1.51color(red)(cancel(color(black)("s"))) = "14.8 m/s"`

The book will have maximum velocity after 1.51 seconds of free falling, just before it impacts the ground.