# Vertical displacement? help anyone. =]

1. Sep 8, 2008

### mje537

vertical displacement!??? help anyone. =]

i have no idea wat to do!??

In a scene in an action movie, a stunt man
jumps from the top of one building to the
top of another building 3.9 m away. After a
running start, he leaps at an angle of 13◦ with
respect to the flat roof while traveling at a
speed of 5.9 m/s.
The acceleration of gravity is 9.81 m/s2 .
To determine if he will make it to the other
roof, which is 1.3 m shorter than the build-
ing from which he jumps, find his vertical
displacement upon reaching the front edge of
the lower building with respect to the taller
building. Answer in units of m.

2. Sep 8, 2008

### Redbelly98

Staff Emeritus
Re: vertical displacement!??? help anyone. =]

Welcome to Physics Forums mje537.

For homework questions you must supply the relevant equations and show an attempt at solving the problem, before we can help you.

What equations are used for this type of problem?

3. Sep 8, 2008

### mje537

Re: vertical displacement!??? help anyone. =]

well i tried the time equation.
t=2V0Sindeta/g
and then i used the displacement formula
y= V0t + 1/2gt2
but they told me it was wrong.
i dont want the answer i just want to know the concept of how to get the answer.
wat i do wrong.???

4. Sep 8, 2008

### Redbelly98

Staff Emeritus
Re: vertical displacement!??? help anyone. =]

First try using

x = V0x t + ½ ax t2

to figure out how long it takes to reach the front edge of the lower building.

Note: you'll have to think about what V0x and ax are.

5. Sep 8, 2008

### mje537

Re: vertical displacement!??? help anyone. =]

to find the time i used.
y=2V0Sindeta/g
y=2(5.9)(sin13)/9.81
y=2.7

then i used ur equation.

x=V0x t + ½ ax t2
x=(5.9)(2.7)+1/2(9.8)(2.7)2
x=15.93+35.721
x=51.651

so did i do anythin wrong.???

6. Sep 8, 2008

### Redbelly98

Staff Emeritus
Re: vertical displacement!??? help anyone. =]

This equation looks wrong to me.

I'm suggesting you begin with my equation. It describes the motion in the horizontal (x) direction. You need to know:
1. The acceleration in the horizontal direction (ax). Hint: it is not g.
2. The initial speed in the horizontal direction.

7. Sep 8, 2008

### mje537

Re: vertical displacement!??? help anyone. =]

im sorry but phyiscs is my most complicated subject and i dont know
wat you mean by a.

is it the acceleration.

8. Sep 8, 2008

### mje537

Re: vertical displacement!??? help anyone. =]

y=(5.9)(3.9)+1/2(1.3)(3.92)
y=23.01+9.8865
y=32.896

9. Sep 9, 2008

### Redbelly98

Staff Emeritus
Re: vertical displacement!??? help anyone. =]

Here is the meaning of the terms in my equation:

x is the horizontal displacement
v[wub]0x[/sub] is the x-component of the initial velocity
t is time
ax is the horizontal component of the acceleration

It seems you have not yet learned a number of concepts that are required for solving problems like this. Not to criticize you, but you may want to schedule time with your professor or teacher, or a tutor. Somebody who can help you with this stuff in person.

The concepts needed for solving this problem:

Treating motion in the vertical and horizontal directions separately.
Using trigonometry to come up with the horizontal and vertical components of velocity (or any vector)
Knowing, for example, that "t" is used for time in these equations. This is a pretty widespread convention, I've never known anybody who used "y" for time.
Knowing that gravity provides a downward acceleration, therefore does not affect horizontal motion.

10. Sep 9, 2008

### Redbelly98

Staff Emeritus
Re: vertical displacement!??? help anyone. =]

Since ax is zero (no horizontal acceleration), my earlier equation actually simplifies to

x = v0 cos(angle) t

Plug in quantities from the given information, and solve for t = time taken to reach the other building's front wall. That is a start to finding the solution, but not the final solution.

Also: I don't know if this would help, but here is some online info that covers this material:
http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html

For this problem in particular, scroll about 1/4 down the page to "General Ballistic Trajectory".

And please do consider talking to your teacher one-to-one ... the earlier you do that the better.

Good luck,

Mark