How to calculate time when falling off a building

AI Thread Summary
To calculate the time it takes for an object to fall from a height of 25 meters, one can use the formula h = (gt^2)/2, where g is the acceleration due to gravity (approximately 9.81 m/s²). The discussion highlights the need for understanding kinematic equations, which relate distance, speed, and acceleration, particularly when speed is not constant. Basic algebra and physics knowledge are sufficient for this problem, but more complex scenarios involving variable acceleration may require calculus. The conversation emphasizes the importance of grasping these foundational concepts in classical mechanics for solving motion-related problems.
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Homework Statement



A friend of mine and me were walking through the city when he looked up at a tall building, let's say 25 metres, and said "I wonder how long it'd take to hit the ground if you jumped off". I was thinking about it and realized I didn't even know what the appropriate technique(s) are to solve such a problem.

So: where should I begin if I have an object and a height, (on Earth), and I want to know roughly how long that object would take to hit the ground when dropped from that height?

The Attempt at a Solution



I first thought about the old speed = distance/time formula. Obviously I want not speed but time as the subject so I pondered time = distance/speed. But of course the speed would be changing all the time due to acceleration from gravity. This made me think I should be using calculus. But I don't really know any apart from bare basic differentiation.

So I suppose I could figure out something like, 'the object has fallen 9.8m after 1 sec, then 9.8 + 9.8(2)m after 2 secs' etc. but if that's even correct, surely there's a better way - what do I do if I end up with a distance greater than the height of the building?

So then I thought I needed to integrate something but I have only done indefinite integrals which I think gives you a function from which you differentiate to get that which you just integrated. But I have no clue either what that function should look like, or what function I should be integrating, or if my friend's question can even be answered with basic calculus.

Would I have to compute something like, time = distance/9.8ms-2 ?

Thankyou.
 
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Ignorimng air resistane,
h = (gt^2)/2 & solve for t. g = your 9.81m/sec^2. h in m, t in sec.
 
Thankyou very much. Please can you tell me which area of mathematics I would learn about such things i.e. speeds and distances and times of moving things when the speed is not constant. So obviously beyond the basic s = d/t stuff. Would that be in classical mechanics?
 
there is a set of equations relating s, u, v, a and t. they are called the kinematic equations. They should be covered in last couple of years of high school/6th form (years 12 & 13)
 
Oh cool, thanks! Yeah I am afraid I stopped at year 11.

s, u, v, a and t.

What is u? And is 'a' acceleration?

Thanks.
 
3.141592 said:
Thankyou very much. Please can you tell me which area of mathematics I would learn about such things i.e. speeds and distances and times of moving things when the speed is not constant. So obviously beyond the basic s = d/t stuff. Would that be in classical mechanics?

This question is high school algebra or physics.

If acceleration (g) were not constant then you would have to have covered elementary calculus (1st semester college) and probably 1st semester introductory engineering-level college physics. Example: a car puts out constant power. What is the relationship between distance covered s and time t if s(t=0) = 0? This requires both calculus and probably 1st sem. eng level physics also. The calculus should probably include elementary differential equation solution.

(Ans. s = (2/3)(2P/3m)1/2t(3/2) where P = power, m = mass of car. IF I did it right!)
 
3.141592 said:
Oh cool, thanks! Yeah I am afraid I stopped at year 11.

s, u, v, a and t.

What is u? And is 'a' acceleration?

Thanks.

u = initial speed
s = displacement
v = final speed
t = time
a = acceleration
 
Thankyou both.
 
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