Finding the Height of a Building Using Free Fall Motion

[Zero_Bishop]

I'm really bad at physics and never have time for study groups and was wondering if you guys/gals might be able to help me out with my problem.

If a person steps off a building of height h, and free falls on the way down to the bottom. What is the height of the building if he falls a distance of h/4 in his last 1 second of fall.

Any ideas?

 

[camherokid]

Hight = 4.9 * t^2.
Free fall, time dependable..

 

[ganstaman]

I don't know how much help we're supposed to give for these questions here, but here's the general method I would employ if I were solving this:

Get out a piece of paper and draw the situation, labeling distances, times, velocities, and accelerations. If you don't do this, you may find that you don't use consistent labels throughout the problem and everything will get messed up.

Next, draw a table with 3 columns. In the first column, write all values that you know (essentially constants in the scope of this problem). For you, these would be the 1 second of fall (whatever you decided to label that time period), and g=9.8m/s^2 (standard gravity, it gets written in this column for nearly every question). In the second column, you write what you need to find out: h in this instance.

For the last column, write every equation that you know that could potentially link these values (the 'what you know' and 'what you want to find'). Basically, these four here: http://en.wikipedia.org/wiki/Kinematics#Equations_of_uniformly_accelerated_motion

Then, start figuring out which equation you'd like to use and plug in the variables you created. Put enough of it together and most variables should drop out.

Here, you would know h if you knew h/4. You know that the person traveled a distance of h/4 in time=1 second at acceleration=g=9/8m/s^2 at an initial velocity of some variable. You could put all this into an equation and solve for h, but you still have another variable. Use more equations and everything and it will work itself out.

 

[FedEx]

I'm really bad at physics and never have time for study groups and was wondering if you guys/gals might be able to help me out with my problem.

If a person steps off a building of height h, and free falls on the way down to the bottom. What is the height of the building if he falls a distance of h/4 in his last 1 second of fall.

Any ideas?

Hi
I have to accept that it is quite a good problem for starters. Now I have got the answer but as per the rules i cannot give you the direct answers. So first of all tell me,do you know the basic equations of kinematics? like v^2-u^2=2ad, where v is the final velocity and u is the initial velocity and d is the displacement.

 

[Killer slug]

is there and wind involved and is our friend airodynamic or flailing

 

[Zero_Bishop]

This is set in ideal settings, man stops off of the top of a building no upward motion, no air resistance to be calculated for at all. Yes, I do know the kinematic motion equations but was wondering how to use them seeing as to what little knowledge I have. Remember I do not know his final velocity or time for the man to fall the other 3/4 height of the building. Thank you all for you help thus far.

 

[Saladsamurai]

This is set in ideal settings, man stops off of the top of a building no upward motion, no air resistance to be calculated for at all. Yes, I do know the kinematic motion equations but was wondering how to use them seeing as to what little knowledge I have. Remember I do not know his final velocity or time for the man to fall the other 3/4 height of the building. Thank you all for you help thus far.

You have initial velocity, a displacement, and a time, and acceleration.

Can you write the equation that relates these variables?

Casey

 

[ganstaman]

This is set in ideal settings, man stops off of the top of a building no upward motion, no air resistance to be calculated for at all. Yes, I do know the kinematic motion equations but was wondering how to use them seeing as to what little knowledge I have. Remember I do not know his final velocity or time for the man to fall the other 3/4 height of the building. Thank you all for you help thus far.

What I wrote last left you with an equation to fill in that had 2 unknowns: h and v (with v being the velocity after falling (3/4)*h, in other words, the velocity at the start of the last quarter of the fall).

Obviously, you can't just find h here, so you have to find something else for v. Ask yourself why this man is traveling at velocity v. That is, at the beginning of the problem he is clearly not traveling in the verical direction, but now he is. Is there any equation you could write that would incorporate this velocity he now has? Hint: if you do it right, you should have 2 unknowns in this equation as well, v and h.

And so with two equations with the same two unknowns, all is easily solvable.

What have you tried so far? Which equations were you attempting to use but found they didn't work? Have you just not known which equations to use? Remember, with acceleration and velocity, there are equations that either use time, displacement, or both. So you often need less information than you'd think.

 

[FedEx]

This is set in ideal settings, man stops off of the top of a building no upward motion, no air resistance to be calculated for at all. Yes, I do know the kinematic motion equations but was wondering how to use them seeing as to what little knowledge I have. Remember I do not know his final velocity or time for the man to fall the other 3/4 height of the building. Thank you all for you help thus far.

Hi
Now see time is given so we can use d=ut+1/2at^2. Here d is h/4 and t is 1 and a is g. So you get the initial velocity of the body of the last one second. Now you can use v^2-u^2=2ad for the last one second so d is equal to h/4 and a=g and v is the final velocity for the last one second.
Now here for u we can replace the value which we got by the previous equation.Now we apply v^2-u^2=2ad for the whole free fall. So d=h and a=g and u=o. So you will get an equation in terms of v^2=2gh.This is the third equation.
Now we have got the second equation also in terms of v^2 and h and g.As we have replaced u by the value which we were getting in the first equation.So solve the second and third equation. You will get h.