Solving Projectile Motion with Initial Height and Final Velocity

[Tubs]

I've done these questions a million times, but it was last year and I'm stumped. if you have:

velocity of projectile
angle
range

and the initial height is greater than that of the final height, how do you solve for initial height and final velocity?

 

[Seiya]

the law of conservation of energy is much better :P

 

[ringojuice]

I have a similar issue.. I took AP Phys B last year and this year I can't remember the simplest things!
I am completely stumped on a problem whose only known variables are the maximum height of the trajectory and the horizontal distance the object travels.
I can't seem to figure out a way to find the velocities involved!
Help...

 

[Rake-MC]

Perhaps giving us exact examples would help, also showing what you have tried is always a good start.

 

[ringojuice]

The problem is really simple. It involves a guy throwing a pebble at a window such that when it hits the window at a height of 4.5 m, it has only horizontal velocity (it has reaches the top of its trajectory). The distance to the window horizontally is 5.0 m, and I am asked for the horizontal velocity.
No matter which equation I try, I always seem to have more than one unknown; for example, the position equation x=initial position + v(initial)t + .5at^2. I'd need v to solve for t and vice versa.
The only thing that I thought might work was neglecting the window and assuming a complete trajectory, but I still end up with more than one variable.
the more I look at this problem, the more muddled I become :(

 

[ringojuice]

I also tried calculating the angle of the trajectory based on the distances given and came up with approximately 41 degrees... but from there I didn't know how to calculate the vector components of the velocity because no velocities are given.

 

[Rake-MC]

So you're solving for the height it's thrown from correct? You don't have final velocity or initial velocity? Well if you ask me, you're in a bit of a pickle..

 

[ringojuice]

no problem, thanks for your attention to this!

 

[ringojuice]

no, I'm solving for the horizontal velocity of the pebble. I know the height of the trajectory, so the height from which it is thrown is irrelevant, I think. I am also given the horizontal distance to the window.
Basically, my issue is that I can't figure out how to isolate and solve for any of the velocities.

 

[Rake-MC]

What do you mean you know the height of the trajectory..? The trajectory under uniform, one dimensional, constant acceleration is parabolic. The trajectory is not a constant value.

If you knew the height it was thrown from, we could solve it.

 

[ringojuice]

sorry, I meant the maximum height that the pebble reaches relative to the point from which it was thrown..

 

[Rake-MC]

Okay, now we're getting somewhere. How far (horizontally and vertically) is this from its final point?

 

[ringojuice]

It is 5.0 m from the window horizontally, and 4.5 m away vertically.

 

[Rake-MC]

Okay, this is good. I now realize that you mentioned this earlier. My apologies for not picking up on it then.

First we want to find the time that it takes for the projectile to fall from this peak point to the same vertical height as the window (we'll label this point zero).
From this, we have:

Acceleration = 9.8ms^-2
Displacement(y) = 4.5m
initial velocity (y) = 0ms^-1
you can now solve for the time it takes to fall the distance, I'll give you a head start.

s = ut + \frac{1}{2}at^2

where u = initial velocity
t = time
a = acceleration
s = displacement

Let me know if you're understanding this (I don't have a good gauge of how advanced you are yet)

 

[ringojuice]

I understand this explanation, but I fear I've miscommunicated the problem :(
4.5 meters is both the maximum height that the pebble reaches as well as the height at which it hits the window.

 

[Rake-MC]

Okay, that makes me very confused then. When it is 5m horizontally from the window, it is 4.5m vertically from the ground. And when it is 0m horizontally from the window (when it hits it) it's is also 4.5m vertically from the window.

And yet 4.5 is the highest point in the pebbles trajectory? This doesn't really make sense to me sorry. Is there a diagram we can refer to?

 

[ringojuice]

let's see... there's no diagram that I know of..
I'm assuming the point from which it was thrown as "ground level" or zero vertical distance. I figure the height of the man throwing the pebble is irrelevant to its motion.
That way, the pebble hits the window at 4.5 m from its starting point at zero, and the horizontal distance remains 5.0 m.

 

[Rake-MC]

http://img184.imageshack.us/img184/5488/diagramce3.png

Okay I hope we've got this right this time!

So that's all the information we're given, are we specifically told that it hits the window at the peak of its trajectory?

We can also assume that gravity is 9.8ms^-2

If all of this is true, then we can solve the equation.

 

[ringojuice]

yes! I think we've got it this time.
Thank you so much for spending this time!

 

[Rake-MC]

Just realized that I got 5m and 4.5 m backwards haha.

Okay so first we work with vertical axis.
We have:

Acceleration (9.8ms^-2)
Final velocity (0ms^-1)
displacement (4.5m)
solving for initial velocity (am I right?)

so first for y velocity:

v^2 = u^2 + 2as
where v = final velocity
u = initial velocity
a = acceleration
s = displacement.

Let me know how you go with this.

 

[ringojuice]

right! I've gone through this step before, and I got 4.6981 for v.
then I plugged that into the y displacement equation to solve for time and got an unsolvable quadratic:
4.905t^2 - 4.6981t +4.5 = 0

 

[Rake-MC]

Hold on, when you say you got 4.6981 for v, do you mean initial or final velocity? We already know final is zero, and when I solve this I get 9.4ms^-1
This is just: u = \sqrt{-(2)(-9.81)(4.5)} Do you understand this?

Now we want to solve for time. We can use: v = u + at.
Are we working on the same wavelength so to speak?

 

[ringojuice]

oh! I think we're onto my mistake here.
I multiplied by .5 instead of by 2.
what a silly mistake!
I think if I solve this from here, I'll be fine.
goodness! haha. thank you ever so much for your help! :) and I'm sorry I took your time with such a silly mistake!

 

[Rake-MC]

Don't worry about it, we're here to help you haha! Don't hesitate to post again if you still can't solve it. Sorry that it took so long to understand your question.. I'm terrible at questions without a diagram infront of me..

 

[ringojuice]

But really, thanks so much. What a lifesaver! I would have been stuck on this for a long time... this is what sleep loss does. I will probably be back on the forums often!
Thanks again!