Question

Can someone help me solve this problem? Thank you very much! 11th grade level.

Answer 1

A) `v_(fy) = 14.7 => -"15 m/s"`

B) `Deltay = -11.025 => -"11 m"`

Explanation:

You'll need to use two important kinematic equations to solve this problem:

`Deltay = v_(0y)t + 1/2a_yt^2`

`2a_yDeltay = v_(fy)^2 - v_(0y)^2`

Do not be put off by all the subscripts -- the `y`'s merely signify that this problem deals with vertical motion (i.e. on the `y`-axis) as opposed to horizontal. So, let's see how these equations will apply:

I'm going to do part B first, since we can use the information from B to solve part A:

B) Let's paint a picture of what's happening here: we have an object that is dropped from a certain height. Let's make some inferences:

• The fact that the book is " dropped " means that it has no initial velocity. Sure, it will pick up speed, but at the moment you let go, the book has no velocity.

Note: This would be different if you `color(red)(threw)` the book down with some speed, but if you just drop it you can assume no initial velocity.

• There will be an acceleration caused by gravity , pulling the book downwards. Hence, you will need to take this into account when setting up your equation.

Now, I'm going to rewrite the first kinematic equation from above based on what I know:

`Deltay = cancel(v_(0y)t) + 1/2a_yt^2`

`=> Deltay = - 1/2"g"t^2`

Where `g = 9.8 m/s^2`

Note: there is a negative sign in front of g because we've defined up as the positive direction. You could just as easily set down as your positive direction, and make that term positive. As long as you're consistent through out, it doesn't matter

Now we just plug in and solve:

`=> Deltay = - 1/2(9.8)(1.5)^2`

`=> Deltay = -11.025 => -"11 m"`

rounded to 2 significant figures.

A) For part A we'll use the second equation. Before we plug in numbers let's rearrange it to solve for final velocity:

`2a_yDeltay = v_(fy)^2 - v_(0y)^2`

`=> v_(fy) = sqrt(2a_yDeltay + cancel(v_(0y)^2))`

`=> v_(fy) = sqrt(2(-g)Deltay)`

Now you just plug in what we know:

`=> v_(fy) = sqrt(2(-9.8)(-11.025)`

`v_(fy) = 14.7 => -"15 m/s"`

rounded to 2 significant figures

There you are. I'm sorry this answer turned out kind of long, but I wanted you to see the thought process behind how to set up and solve these kinds of problems. This is super important to understand, especially once you get to more advanced physics.

Hope that helped :)