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A falling object's speed

  • Thread starter phuketman
  • Start date
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1. Homework Statement

A 69 kilogram person falls from a 35 meter tower head first to the ground

2. Homework Equations

What is the maximum velicoity he reaches before impact?

3. The Attempt at a Solution
 

DaveC426913

Gold Member
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See that #3? That's where you make an attempt at a solution.

We don't spoon-feed homework here.
 
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Hint: Use your kinematic equations for uniform acceleration.
 
3
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Sorry if I am a dumbum. I'm 57 and have forgotten all my equations, not that I was very good at physics. I am doing research for a stuntman who has survived this fall and want to know what speed he reached.
 
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DaveC426913

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The https://www.physicsforums.com/showthread.php?t=5374" that you agreed to when you signed up require you to attempt a solution:

"NOTE: You MUST show that you have attempted to answer your question in order to receive help."


Furthermore, we are expressly requested not to do work for people here; this is afterall, and homework help forum.
 
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DaveC426913

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V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

V is final velocity in units of meters per seconds (m/s)
Vo is initial velocity in units of meters per seconds (m/s)
a is acceleration in units of meters per squared seconds (m/s2)
t is time in seconds (s)
X is final displacement in units of meters (m)
Xo is initial displacement in units of meters (m)
 

Integral

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Science Advisor
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The mass does not matter, any falling object will have the same speed after falling 35m.

First you need the time of the fall.

x = fall distance = 35m
a = acceleration due to gravity = 9.8 m/s
t = time of fall.

x = .5 a t^2
[tex] t = \sqrt { \frac {2x} a } [/tex]

Given the time of the fall the speed is

v = at

so [tex] v = \sqrt {2 a x} [/tex]

That should give you all the equations you need. can you take it from here?
 
3
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thank you DaveC426913 and Integral. I am new to this board and appreciate your help.
I will not bother you again.
 

DaveC426913

Gold Member
18,326
1,920
thank you DaveC426913 and Integral. I am new to this board and appreciate your help.
I will not bother you again.
Feel free to bother. That's what we're here for. :wink: But there's rules is all.
 

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