How to account for gravity on a given force?

[SpunkyMonkey]

Suppose a force exerted horizontally on an object displaced it a total of 100 meters horizontally over 4 seconds.

Suppose that we want to estimate how far that same force would displace that object if it was instead vertically directed straight up against gravity.

Would this be a valid way to account for the effect of gravity on that upward displacement:

upward displacement = 100 - 9.8(100/4²) = 39 m​

So the same force that when directed horizontally displaced the object 100 meters horizontally if instead directed vertically (straight up) on that object would displace it approximately 39 meters vertically (assuming all other factors equal). No?

Btw, this is not homework and I'm not a student.

 

[willem2]

It is correct to subtract the amount the object would fall in 4 seconds from the upward displacement of 100 m due to the force. The amount the object falls in 4 second is \frac {1}{2} g t^2 with g = 9.8 m/s^2

You should really try to write down more of your reasoning, I have no idea where 9.8\frac {100}{4^2} comes from

 

[cepheid]

I don't think your equation is quite right. The upward displacement in this case will just be the horizontal displacement (100 m) MINUS the amount the object would have fallen if gravity alone were acting on it. This result is not obvious, but it is easy enough to derive.

You can figure out the net force that must have acted on the object (horizontally). You know that for motion under constant acceleration:

\Delta x = v_0 \Delta t + \frac{1}{2}a (\Delta t)^2

We know that Δx = 100 m and Δt = 4 s. Therefore, assuming that it started from rest (v₀ = 0), we can solve for the acceleration:

a = \frac{2 \Delta x}{(\Delta t)^2}

We know from Newton's 2nd law that the net horizontal force on the object must therefore have been ma, with a given by the expression above.

Now, if this force ma acts vertically on the object, and gravity mg acts vertically downward, then the net upward force is ma - mg = m(a - g). So the upward acceleration on the object is a - g. (It just so happens that for the numbers you have chosen, a > g, and the object will move upward). We can compute the vertical displacement Δy using the same kinematic formula for motion under constant acceleration (only this time the acceleration is equal to a - g):

\Delta y = v_0 \Delta t + \frac{1}{2}(a - g) (\Delta t)^2

Again, assuming we start from rest, this becomes:

\Delta y =\frac{1}{2}\left(\frac{2 \Delta x}{(\Delta t)^2} - g\right) (\Delta t)^2

= \Delta x - \frac{1}{2}g(\Delta t)^2

Now, since (1/2)g(\Delta t)^2 is just the amount that the object would have moved downward under free fall, this result confirms what I said at the very beginning of my post that the upward displacement is just 100 m minus the amount the object would have fallen in 4 s under free-fall.

 

[SpunkyMonkey]

Thanks cepheid and willem2!

So if I understand, in this case the amount the object would fall (neglecting air resistance) in 4 seconds would be

0.5(-9.8)(4²) = -78.4 meters

And so the same force exerted vertically would displace the object approx 100 - 78 = 22 meters upward.

 

[cepheid]

Thanks cepheid and willem2!

So if I understand, in this case the amount the object would fall (neglecting air resistance) in 4 seconds would be

0.5(-9.8)(4²) = -78.4 meters

And so the same force exerted vertically would displace the object approx 100 - 78 = 22 meters upward.

Yeah, that sounds about right.