What is the Kinematics of a Particle Leaving a Ledge?

[jjones1573]

Homework Statement

A particle leaves a ledge at a height of 4m and hits the ground at 1.2s later landing 2m away horizontally.

find:
The original velocity
The final velocity
The max height acheived

Homework Equations

So I know that for the original velocity:

Vo = D * SQRT(g/2H)

where:
Vo = original velocity
D = how far it landed
g = gravity
H = original height

The Attempt at a Solution

So for my initial velocity with the above equation I get:

Vo = 2 * SQRT( 9.8 / 2 * 4)

= 2.1m/s

Is this method correct?

I'm not sure how to find the final velocity before it hits the ground

Also the max height is surely the original height of 4m as its going down from here. suppose the particle jumped from its position on the ledge could I figure out a max height.

One last thing. could anybody show me how to find an acceleration vector of a projectile. I know how to get the velocity vector using arctangent but I can't find anything on acceleration vectors.

thanks so much,

 

[The Tutor]

A particle leaves a ledge at a height of 4m and hits the ground at 1.2s later landing 2m away horizontally.

find:
The original velocity
The final velocity
The max height achieved

Well I'm presuming the particle is experiencing horizontal projectile motion.To find the original velocity I would use dx=vx+1/2a(t)², where x are the horizontal problems given in the question, such as the particle landing 2 m away horizontally, time can be used in both x and y component equation.

The final velocity would be the velocity upon impact.. now you can use v_2y²=v_1y²+2ad. Because of the horizontal velocity, v_1y would equal zero, simply plug in the other numbers.

You can easily find the maximum height after finding the first two parts, try it yourself :P.

 

[ashishsinghal]

Well I'm presuming the particle is experiencing horizontal projectile motion.

No, that is wrong.

 

[ashishsinghal]

So I know that for the original velocity:

Vo = D * SQRT(g/2H)

where:
Vo = original velocity
D = how far it landed
g = gravity
H = original height

For which situation is this true? Doesn't it depend on the angle on which it is thrown?

 

[The Tutor]

For which situation is this true? Doesn't it depend on the angle on which it is thrown?

I said that because there wasn't an angle being given.

 

[ashishsinghal]

Enough information is given to find the angle

 

[jjones1573]

Oh I had found that equation on a thread on yahoo answers where someone had posed a similar question and that was the answer given. So how could I find the angle or is there some resource you could point me to where I could find this?

 

[ashishsinghal]

Let the velocity be u and angle be x.
ucosx.t = 2
and
1/2gt^2 - usinxt = 4

From here use t=1.2 to get usinx, ucosx.

 

[jjones1573]

sorry I'm kind of confused here. Do I need to find the initial velocity first in order to do this?

 

[ashishsinghal]

No, you don't. Obtain usinx and ucosx as above. Divide them to get tanx. You have got the angle in tan

 

[jjones1573]

Ah ok I get it thanks so much!

so to obtain initial velocity do I just find the components. when I get usinx and ucosx I divide them by sinx and cosx and then calculate the intial velocity from these x and y components

Also when calculating the max height I use Viy * t + (0.5)at^2 I use this equation when the projectiles height remains the same at the intial and final stages but in this instance should I add my initial height to the answer or do I need to do somehting else?