A falling object travels one-fourth of its total distance in the last second of its fall....

In summary: You should try to derive this equation. I got it by solving for ##D## directly. Alternatively, and probably better, if you use @Ray Vickson's...
  • #1
6
1

Homework Statement


"A falling object travels one-fourth of its total distance in the last second of its fall. From what height was it dropped?"

Reference: Up is (+)

Scenario 1 (Complete motion)
a = -9.8
v1 = 0
d = -x
v2 = ?

Scenario 2 (Motion at the last second)
a = -9.8
d' = (1/4)(-x)
t' = 1
v2 = ?

Homework Equations


Complete motion: v22 = v12 + 2ad
Motion at the last second: d' = v2t' - (1/2)(a)(t')2

3. The Attempt at a Solution

Setting both equation equal to v2 and then solving for x

Complete motion:
v22 = v12 + 2ad
v22 = (0)2 + (2)(-9.8)(-x)
v22 = (19.6)(x)
v2 = [(19.6)(x)](1/2)

Last second:
d' = v2t' - (1/2)(a)(t')2
(1/4)(-x) = v2 (1) - (-4.9)(1)2

Substitution:
(1/4)(-x) = [(19.6)(x)](1/2) - 4.9
[(1/4)(-x)]2 = (19.6)(x) + (-4.9)2
x2 / 16 = (19.6)(x) + (-4.9)2
x2 - (313.6)(x) - 384.16 = 0

x = - 1.22 x = 314.82

Edit: Was a problem with the signs should be
(1/4)(-x) = [(19.6)(x)](1/2) + 4.9

Answer is supposed to be x = 270
 
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  • #2
spacey__ said:
d' = v2t' - (1/2)(a)(t')2
(1/4)(-x) = v2 (1) - (-4.9)(1)2

Why “-“ in front of “1/2 a t^2”? If that represents the downward direction then you shouldn’t substitute in 4.9 with a minus sign. You can put the direction explicitly into the equation of motion, or you can substitute in numbers with signs, but you can’t do both. Those two negatives in front of 4.9 are the same negative put in twice.
 
  • #3
spacey__ said:
v22 = (0)2 + (2)(-9.8)(-x)

In the “last second” section you have v2 as the speed at the BEGINNING of the last second. Here you are using v2 as the speed at the end of the motion. You need to be consistent.
 
  • #4
This would probably go better if you made a nice diagram with axes and labels showing where the coordinates are zero, which direction is positive and most importantly which locations you mean by x1, x2, v1, v2
 
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  • #5
Thanks for the reply!

Regarding the negative in the equation "d' = v2t' - (1/2)(a)(t')2" I got it off my high school notes which should not include direction within the equation itself. However I've checked this below by changing the frame of reference so that down is (+).

The equation that you mentioned, "v22 = (0)2 + (2)(-9.8)(-x)" is apart of the full motion in which v1 represents the the ball dropping at rest and v2 hitting the ground. I've included a diagram to better represent these variables.

The reworked solution is here: https://imgur.com/0uLwxyS

My answer is 272.99 when it should be 270.
 
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  • #6
spacey__ said:

Homework Statement


"A falling object travels one-fourth of its total distance in the last second of its fall. From what height was it dropped?"

Reference: Up is (+)

Scenario 1 (Complete motion)
a = -9.8
v1 = 0
d = -x
v2 = ?

Scenario 2 (Motion at the last second)
a = -9.8
d' = (1/4)(-x)
t' = 1
v2 = ?

Homework Equations


Complete motion: v22 = v12 + 2ad
Motion at the last second: d' = v2t' - (1/2)(a)(t')2

3. The Attempt at a Solution

Setting both equation equal to v2 and then solving for x

Complete motion:
v22 = v12 + 2ad
v22 = (0)2 + (2)(-9.8)(-x)
v22 = (19.6)(x)
v2 = [(19.6)(x)](1/2)

Last second:
d' = v2t' - (1/2)(a)(t')2
(1/4)(-x) = v2 (1) - (-4.9)(1)2

Substitution:
(1/4)(-x) = [(19.6)(x)](1/2) - 4.9
[(1/4)(-x)]2 = (19.6)(x) + (-4.9)2
x2 / 16 = (19.6)(x) + (-4.9)2
x2 - (313.6)(x) - 384.16 = 0

x = - 1.22 x = 314.82

Edit: Was a problem with the signs should be
(1/4)(-x) = [(19.6)(x)](1/2) + 4.9

Answer is supposed to be x = 270

A much easier way is to let distance down be positive, so for a trip time of ##T## we must have
$$g(T-1)^2 = \frac{3}{4} g T^2, \; T > 1$$
This is a simple quadratic equation having two positive roots, but only one of which is ##> 1.## The drop height is ##g T^2.##
 
  • #7
Ray Vickson said:
The drop height is ##g T^2.##

Ray, do you mean ##{g \over 2} T^2##? Then the answer is once again 272.99.
 
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  • #8
@spacey__ A better answer is:

##x = \frac{2g}{7 - 4\sqrt{3}}s^2##

Then it's clear what is a rounding error.
 
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  • #9
verty said:
Ray, do you mean ##{g \over 2} T^2##? Then the answer is once again 272.99.

Yes, of course: ##H = (1/2)g T^2.##
 
  • #10
PeroK said:
@spacey__ A better answer is:

##x = \frac{2g}{7 - 4\sqrt{3}}s^2##

Then it's clear what is a rounding error.

So the textbook answer was a rounding error? What is s in this equation and where is this equation from?
 
  • #11
Ray Vickson said:
Yes, of course: ##H = (1/2)g T^2.##

The total trip time is unknown in the complete motion. How should I use this equation?
 
  • #12
spacey__ said:
The total trip time is unknown in the complete motion. How should I use this equation?
NO: after solving the quadratic equation for ##T## it is not unknown anymore!
 
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  • #13
spacey__ said:
So the textbook answer was a rounding error? What is s in this equation and where is this equation from?

##s## is seconds. Strictly speaking, this is needed to keep the units correct.

You should try to derive this equation. I got it by solving for ##D## directly. Alternatively, and probably better, if you use @Ray Vickson's approach and solve for ##T## first, you get the equivalent:

##x = 2(7 + 4\sqrt{3})gs^2##

Note that if you drop the ##s^2##, then you have a distance equal to an acceleration, which is dimensionally wrong. The ##s^2 = (seconds)^2## sorts this out.
 
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1. How is distance calculated for a falling object?

The distance of a falling object is calculated using the formula d = 1/2 gt^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

2. What is the total distance traveled by a falling object?

The total distance traveled by a falling object can be calculated using the formula d = 1/2 gt^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the total time of the fall.

3. Why does a falling object travel one-fourth of its total distance in the last second of its fall?

This is due to the acceleration due to gravity being constant. As the object falls, its speed increases and it covers more distance in each second. In the last second, the speed is at its highest and thus the object travels one-fourth of the total distance.

4. Does the mass of the object affect its total distance traveled?

Yes, the mass of the object does affect its total distance traveled. Heavier objects will fall faster and thus cover more distance in a shorter amount of time compared to lighter objects.

5. Is air resistance a factor in calculating the distance traveled by a falling object?

Yes, air resistance can affect the distance traveled by a falling object. It can slow down the object and decrease its final speed, resulting in a shorter distance traveled. However, for objects falling in a vacuum, air resistance does not play a role in the distance traveled.

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