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1-D Kinematics

  • Thread starter Leoragon
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  • #1
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Homework Statement



Object dropped from a tower at height of 71.3 metres

What is the time it takes to hit the ground? What is the velocity when it does?

gravitational acceleration = 9.8m/s2


Homework Equations



V = ΔX/Δt

X = average velocity X time

The Attempt at a Solution



I don't know a formula that would enable me to get the time.
 

Answers and Replies

  • #2
993
13
Use the WELL-KNOWN equation i.e.

x = ut + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex].
 
  • #3
15
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Homework Statement



Object dropped from a tower at height of 71.3 metres

What is the time it takes to hit the ground? What is the velocity when it does?

gravitational acceleration = 9.8m/s2


Homework Equations



V = ΔX/Δt

X = average velocity X time

The Attempt at a Solution



I don't know a formula that would enable me to get the time.
Try to derive it from formula said by grzz.
u is initial velocity which is zero as they mentioned it that they dropped.
 
  • #4
380
7
Or you can think it in terms of energy mgh=½mv^2 ...etc to solve for the final velocity. On the other hand vy=g*t so you get t.
 
  • #5
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I've tried solving for time and this was what I got:

X= Xo + Vot + 1/2at2

Vo = 0 (it's at rest)

Xo = 0 (also at rest)

a = 9.8m/s2

X = 71.3m

71.3 = 0 + 0 + (.5 X 9.8 X t2)

71.3 = .5 X 9.8 X t2

71.3 / .5 = 9.8 X t2

71.3 / .5 / 9.8 = t2

√(71.3 / .5 / 9.8) = t

t ≈ 3.81s

Does this mean that time equals: √X/.5/a ? If initial velocity and initial position equals 0?

If so, how do you calculate it if the initial velocity and position doesn't equal 0?
 
Last edited:
  • #6
haruspex
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X= Xo + Vot + 1/2at2
how do you calculate it if the initial velocity and position doesn't equal 0?
Same equation. Plug in the known initial position and velocity and solve the quadratic.
 
  • #7
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Same equation. Plug in the known initial position and velocity and solve the quadratic.
Not what I meant. I meant how to solve for "t"
 
  • #8
MarneMath
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Last time I checked, when you drop an object, the initial velocity is 0 and, if you wanted you could make your reference line the point at which you dropped the object and thus have the initial position be 0, but then you have to be careful with your signs.

Is any of this making sense?
 
  • #9
haruspex
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Not what I meant. I meant how to solve for "t"
That's what I thought you meant, and I thought I answered.
Xf= Xo + Vot + 1/2at2
If you know Xf, Xo, Vo and a, plug those in and solve the resulting quadratic to find t. Doesn't matter whether Xo and Vo are zero or nonzero, provided they're known.
 
  • #10
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That's what I thought you meant, and I thought I answered.
Xf= Xo + Vot + 1/2at2
If you know Xf, Xo, Vo and a, plug those in and solve the resulting quadratic to find t. Doesn't matter whether Xo and Vo are zero or nonzero, provided they're known.
But how can you get Xf if you don't have t? And in the quadratic, I'm not sure which ones would be a, b, or c. I'm thinking in the quadratic formula it's something like.

t = x

b = Vi

a = a

c = Xf

so it will then look like: at^2 + Vit - Xf = 0

I'm trying to figure this out.
 
  • #11
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But how can you get Xf if you don't have t? And in the quadratic, I'm not sure which ones would be a, b, or c. I'm thinking in the quadratic formula it's something like.

t = x

b = Vi

a = a

c = Xf

so it will then look like: at^2 + Vit - Xf = 0

I'm trying to figure this out.
Shouldn't the first term be ½at^2?
 
  • #12
haruspex
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But how can you get Xf if you don't have t?
This whole discussion assumes you know everything in the equation except t.
In the OP, you had X0 = 0, V0 = 0, Xf = 71.3, a = g, and you solved it happily. To do that, you effectively rearranged the equation to be
(a/2)t2 + V0t + X0 - Xf = 0
Then you asked what if X0, V0 are nonzero, to which I answered, just plug those nonzero values into the same equation.
 

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