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

[spacey__]

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

 

[Cutter Ketch]

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.

 

[Cutter Ketch]

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.

 

[Cutter Ketch]

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

 

[spacey__]

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.

 

[Ray Vickson]

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 $g T^2.$

 

[verty]

The drop height is $g T^2.$

Ray, do you mean ${g \over 2} T^2$? Then the answer is once again 272.99.

 

[PeroK]

@spacey__ A better answer is:

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

Then it's clear what is a rounding error.

 

[Ray Vickson]

Ray, do you mean ${g \over 2} T^2$? Then the answer is once again 272.99.

Yes, of course: $H = (1/2)g T^2.$

 

[spacey__]

@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?

 

[spacey__]

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?

 

[Ray Vickson]

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!

 

[PeroK]

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.