- #1

kieth89

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Hey, I need some help :( . I've got an exam next Thursday, and I really suck at this stuff right now. So I need to get better! We are studying kinematics (1d and 2d).

In this problem I need help with, we are given the attached diagram and are supposed to derive the following equations from it using the equations listed in part 2. While I've found three of them, I didn't really understand any of them, so will just list them all.

## Homework Statement

Refer to the image for these.

__Part A__

Find the time [itex]t_{H}[/itex] it takes the projectile to reach its maximum height [itex]H[/itex].

Express [itex]t_{H}[/itex] in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex] (the magnitude of the acceleraion due to gravity).

__Part B__

Find [itex]t_R[/itex], the time at which the projectile hits the ground after having traveled through a horizontal distance [itex]R[/itex].

Express the time [itex]t_R[/itex] in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex].

__Part C__

Find [itex]H[/itex], the maximum height attained by the projectile.

Express your answer in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex].

__Part D:__

Find the total distance [itex]R[/itex] (often called the range) traveled in the [itex]x[/itex] direction; see the figure in the problem introduction.

Express the range in terms of [itex]v_{0}[/itex], [itex]\Theta[/itex], and [itex]g[/itex].

## Homework Equations

The 4 basic kinematics equations we have learned:

[itex]v = v_{0} + at[/itex]

[itex]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/itex]

[itex]v^{0}_{2} + 2a(x - x_{0})[/itex]

[itex]\overline{v} = \frac{v + v_{0}}{2}[/itex]

And these:

[itex]v_{y0} = v_{0}sin \Theta [/itex]

[itex]v_{x0} = v_{0}cos \Theta [/itex]

## The Attempt at a Solution

__Part A:__

I'm going to use the equation [itex]v = v_{0} + at[/itex]

Steps...

[itex]v - v_{0} = gt[/itex] (changed a to g for the gravity acting on the projectile)

[itex]t = \frac{Δv}{g} [/itex] (changed the v's to change in v and divided g to isolate t)

And that's it..But it's wrong! The correct answer says that it is [itex]t_{H} = \frac{v_{0}sin \Theta}{g}[/itex]

So, how did they get that? I'm not seeing any errors in my math..what am I doing wrong?

__Part B:__

For this part I want to do the same thing, just with the x component ([itex]v_{0} cos \Theta [/itex]), but that's not right. The right answer is [itex]t_{H} = \frac{2v_{0}sin \Theta}{g}[/itex], which is double the previous answer. This makes sense, because the previous answer said how long it took to get halfway (the highest point is halfway). But it feels like I should be using the x component, not the y component. Can we use the x component, and if not, why, if so - how?

__Part C:__

For this part I use the following equation:

[itex]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/itex]

Steps...

The first thing I do is replace the x with y, since we are finding the max height ([itex]H[/itex])

[itex]y = y_{0} + v_{y0}t + \frac{1}{2}at^{2}[/itex]

Next I substitute the correct equation for [itex]t_{H}[/itex] in place of [itex]t[/itex], which was found in Part A.

[itex]y = y_{0} + v_{y0} \frac{v_{0}sin \Theta}{g} + \frac{1}{2}a (\frac{v_{0}sin \Theta}{g})^{2}[/itex]

Then simplify, and expanding [itex]v_{y0}[/itex], and moving [itex]y_{0}[/itex], also a = g.

[itex]y - y_{0} = v_{0}sin \Theta \frac{v_{0}sin \Theta}{g} + \frac{1}{2}g (\frac{v_{0}sin \Theta}{g})^{2}[/itex]

[itex]Δy = \frac{v_{0}^{2}sin^{2} \Theta}{g} + \frac{v_{0}^{2}sin^{2} \Theta}{2g}[/itex]

And that last equation is the correct answer. Yay! I got 1 out of 3 so far (and understood it).

Alright, one more,

__Part D__

For this one, I use the same equation as last time:

[itex]x = x_{0} + v_{x0}t + \frac{1}{2}at^{2}[/itex]

So, basically I'm doing the same thing as above except [itex]t[/itex] is now going to be the answer to Part B : [itex]t_{H} = \frac{2v_{0}sin \Theta}{g}[/itex] because we are wanting horizontal distance, not max height. We are going to use the x component now as well.

Steps...

[itex]x - x_{0} = (v_{0} cos \Theta)(\frac{2v_{0}sin \Theta}{g}) + \frac{1}{2}g(\frac{2v_{0}sin \Theta}{g})^{2}[/itex]

[itex]Δx = \frac{2v_{0}^{2} cos \Theta sin \Theta}{g} + \frac{1}{2}g\frac{4v_{0}^{2}sin^{2} \Theta}{g^{2}}[/itex]

[itex]Δx = \frac{2V_{0}^{2} cos \Theta sin \Theta}{g} + \frac{2v_{0}^{2} sin^{2} \Theta}{g}[/itex]

[itex]Δx = \frac{2V_{0}^{2} cos \Theta sin \Theta + 2v_{0}^{2} sin^{2} \Theta}{g}[/itex]

And that's all I got. I still don't know the final answer for this part. And I don't feel like I understand (barely) any of it. I don't know when acceleration due to gravity should be negative, or how the answers that are correct were gotten, or even where to start! I have read the 3 chapters in the textbook (Physics, by Giancoli) and took extensive notes on them, but still have no idea! It's really annoying, confusing, and very frustrating. This post took a ton of time as well.

Thanks for taking the time to read all this and any help would be very, very greatly appreciated,

Josh