Projectile question

[exquisitemelody]

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I don't have a specific example, but how would you solve a problem if you're on a cliff and you throw an object in the air with some initial velocity and it falls on the ground below the cliff. How do you find the final velocity? Would it be like if you just threw the ball down with the initial velocity because when the object comes back down, it has the same speed at that point as it did going up, but it's just going down.

 

[MT Headed]

If you are given initial speed, are asked for final speed (i.e. magnitude of the final velocity), and nobody seems to care about the time, then solve it as an energy problem. Energy problems don't use time or direction.

Einit = Efinal
KEinit + PEinit = KEfinal + PEfinal
(1/2) m vinit^2 + m g hinit = (1/2) m vfinal^2 + m g hfinal
(1/2) vinit^2 + g hinit = (1/2) vfinal^2 (((assume that your hfinal is zero)))

this can be rearranged for easier reading:

vfinal^2 - vinit^2 = 2 g hinit

Basically, all of your initial potential energy was converted into kinetic energy, and this is added to your original kinetic energy.

An interesting side effect of this solution is that it doesn't matter if your initial projectile was launched straight up, straight down, sideways, or at some arbitrary angle theta. The result is the same, it will land with the same final speed. Obviously the direction of initial launch will affect (1) when it lands, (2) where it lands, and (3) in what direction it is travelling when it lands. But you didn't ask for those so I'm not going to waste my precious mcat seconds solving for them too.

 

[Arctangent]

As to your first question, I'd probably do some sort of conservation of energy process. And your premise in your second question is correct, assuming no air resistance/other forces. In fact, that's actually a pretty nice and fast to be able to solve for final velocity given that scenario. In that case Vf^2 = Vo^2 + 2ax should work if you are given the height discrepancy (where Vo is the downward velocity as it goes down), but if you are given a time between release and impact and are looking for final speed, you could just do Vf = Vo + gt (where Vo is the initial upwards velocity).

 

[MD Odyssey]

A much cleaner way is to just look at the energies prior to being thrown and assume it all gets converted into kinetic energy.

Initially, you have some potential energy, given by U = mgh and the kinetic energy, which as you've pointed is proportional to the square of the initial velocity, even though it's directed upwards. Ultimately, all that energy will be converted into kinetic energy since the particle is at the bottom of the cliff. So, sum the original kinetic and potential energy, set them equal to the final kinetic energy, and solve for the final velocity. Easy as pie.

Edit: The question the OP asked implied he was only interested in the speed. Now that I've re-read his question, this approach only works if you want to know the magnitude of the velocity. Hopefully, I haven't confused anyone. And JohnWetzel...why didn't you pick up on this and correct me!

 

[Pre-Med Village]

With the initial speed, giving you kinetic energy, and initial potential energy on the cliff, you could solve the problem with conservation of energy depending on how it was phrased.

Or depending on problem phrasing and given information you could follow the projectile motion framework and solve it as a kinematics problem. Picture the initial velocity as the vector sum of horizontal and vertical components. The horizontal component has constant velocity while the vertical component you can just treat as a free fall problem. Find the height of the peak using kinematics, the distance to deplete the initial vertical velocity to 0, then you can find the speed something would have after falling the distance from the peak, starting at Vy = 0 to the ground using kinematics.

 

[neurodoc]

I don't have a specific example, but how would you solve a problem if you're on a cliff and you throw an object in the air with some initial velocity and it falls on the ground below the cliff. How do you find the final velocity? Would it be like if you just threw the ball down with the initial velocity because when the object comes back down, it has the same speed at that point as it did going up, but it's just going down.

Not enough information. At what angle (from the horizontal) and with what speed are you throwing the object while you're standing (at the edge?) of the cliff?

Once the object is released, ignoring air resistance and other real world physics, the only force acting on the object is gravity, and finding the final velocity when the object hits the ground is a simple kinematics problem. You don't need to consider Energy. You just need to know the object's velocity (speed and direction) when released, its height above the ground (where it will land below the cliff) and you can calculate the speed when it hits the ground.

 

[MD Odyssey]

Not enough information. At what angle (from the horizontal) and with what speed are you throwing the object while you're standing (at the edge?) of the cliff?

Once the object is released, ignoring air resistance and other real world physics, the only force acting on the object is gravity, and finding the final velocity when the object hits the ground is a simple kinematics problem. You don't need to consider Energy. You just need to know the object's velocity (speed and direction) when released, its height above the ground (where it will land below the cliff) and you can calculate the speed when it hits the ground.

Yeah, I mistakenly thought the OP was asking for the speed at the ground. There are lots of ways to solve the problem - energy arguments to find the magnitude of the velocity, and then the initial x-component of the velocity lets you find the angle it hits the ground at as well as the y-component of the velocity. Thanks for making me re-read his question.