Intial Velocity of an Object In Free Fall

[KermitFrog]

So I have to calculate the initial velocity of an object I dropped from 2.195 meters that took .96 seconds to hit the ground, given that the Earth's gravitational pull is 9.81m/s^2. What direction/step should I first take in order to solve this? I'm not asking for the answer just a step or two in the direction of finding the answer. Thanks!

 

[MarkFL]

I would begin with:

$$\d{v}{t}=a$$ where $a$ is a constant.

Thus:

$$dv=a\,dt$$

Now, if we integrate with respect to time $t$, we obtain:

$$\int_{v_0}^{v(t)}\,dv=a\int_0^t\,dt$$

And so we have:

$$v(t)-v_0=at\implies v(t)=at+v_0$$

Now, we also know:

$$\d{x}{t}=v(t)=at+v_0$$

Integrating again w.r.t $t$, we get:

$$\int_{0}^{x(t)}\,dx=\int _0^t at+v_0\,dt$$

$$x(t)=\frac{a}{2}t^2+v_0t$$

So, solving for $v_0$, what do you get?

 

[KermitFrog]

I would begin with:

$$\d{v}{t}=a$$ where $a$ is a constant.

Thus:

$$dv=a\,dt$$

Now, if we integrate with respect to time $t$, we obtain:

$$\int_{v_0}^{v(t)}\,dv=a\int_0^t\,dt$$

And so we have:

$$v(t)-v_0=at\implies v(t)=at+v_0$$

Now, we also know:

$$\d{x}{t}=v(t)=at+v_0$$

Integrating again w.r.t $t$, we get:

$$\int_{0}^{x(t)}\,dx=\int _0^t at+v_0\,dt$$

$$x(t)=\frac{a}{2}t^2+v_0t$$

So, solving for $v_0$, what do you get?

v0 = ( x(t)- (a/2)t^2 )/t ?

 

[MarkFL]

v0 = ( x(t)- (a/2)t^2 )/t ?

Yes! (Yes)

I have written this formula equivalently on my "cheat sheet" for kinematics as:

$$v_0=\frac{1}{2t}\left(2\Delta x-at^2\right)$$

So, now we need to identify, from the given data, the following:

$$t,\,\Delta x,\,a$$

And then plug them into our formula...what do you get?

 

[KermitFrog]

Yes! (Yes)

I have written this formula equivalently on my "cheat sheet" for kinematics as:

$$v_0=\frac{1}{2t}\left(2\Delta x-at^2\right)$$

So, now we need to identify, from the given data, the following:

$$t,\,\Delta x,\,a$$

And then plug them into our formula...what do you get?

Is it t= .96, Delta x = 9.81, and a = 2.195?

 

[MarkFL]

Is it t= .96, Delta x = 9.81, and a = 2.195?

Well, you are correct that (don't forget your units) that:

$$t=0.96\text{ s}$$

However, the acceleration $a$ here is that which is due to gravity, so we have:

$$a=g=9.81\frac{\text{m}}{\text{s}^2}$$

and the change in position is:

$$\Delta x=2.195\text{ m}$$

So, plugging those into our formula, what do you find?

 

[KermitFrog]

Well, you are correct that (don't forget your units) that:

$$t=0.96\text{ s}$$

However, the acceleration $a$ here is that which is due to gravity, so we have:

$$a=g=9.81\frac{\text{m}}{\text{s}^2}$$

and the change in position is:

$$\Delta x=2.195\text{ m}$$

So, plugging those into our formula, what do you find?

v(0) = -2.232 m/s ?

 

[MarkFL]

v(0) = -2.232 m/s ?

I get:

$$v_0=\frac{1}{2\left(0.96\text{ s}\right)}\left(2\left(2.195\text{ m}\right)-\left(9.81\frac{\text{m}}{\text{s}^2}\right)\left(0.96\text{ s}\right)^2\right)=-\frac{290681}{120000}\,\frac{\text{m}}{\text{s}}\approx-2.42\,\frac{\text{m}}{\text{s}}$$

Without seeing your computations, I can't say why our answers differ. Do you understand the significance of the negative sign? Which direction are we taking to be positive here?

 

[KermitFrog]

I get:

$$v_0=\frac{1}{2\left(0.96\text{ s}\right)}\left(2\left(2.195\text{ m}\right)-\left(9.81\frac{\text{m}}{\text{s}^2}\right)\left(0.96\text{ s}\right)^2\right)=-\frac{290681}{120000}\,\frac{\text{m}}{\text{s}}\approx-2.42\,\frac{\text{m}}{\text{s}}$$

Without seeing your computations, I can't say why our answers differ. Do you understand the significance of the negative sign? Which direction are we taking to be positive here?

I made an error when plugging the numbers into my calculator! Since gravity pulls objects downward wouldn't the negative sign indicate that the object was going upward?

 

[MarkFL]

I made an error when plugging the numbers into my calculator! Since gravity pulls objects downward wouldn't the negative sign indicate that the object was going upward?

Yes, we have set in our derivation down to be the positive direction, and so a negative initial velocity would mean it was initially moving in the upward direction. :)

Do we need to concern ourselves with the fact that the object then actually moved though a greater distance than the distance from the initial and final positions? It started out at 2.195 m above the ground and then moved up, and then down to the ground...

 

[KermitFrog]

Yes, we have set in our derivation down to be the positive direction, and so a negative initial velocity would mean it was initially moving in the upward direction. :)

Do we need to concern ourselves with the fact that the object then actually moved though a greater distance than the distance from the initial and final positions? It started out at 2.195 m above the ground and then moved up, and then down to the ground...

I don't think so. Thanks for the help!