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1. Homework Statement
I searched around and found a manual, so this time I have the complete question in English (I'll use different numbers though, because mine is the 8th edition, and the manual is from the 14th, but it's not something that affects this exercise):
2.98) An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.50 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?
2. Homework Equations
y = y0 +U0*t + 1/2 * a * t^2 (1)
U = U0 + a * t (2)
3. The Attempt at a Solution
The assumption I made/parameters I took:
 ^ * Positive Direction is upwards.
  (+) * y = 0 is at the bottom of the cliff/the ground.
 * H = 3h
 * H is the whole height of the cliff.
 *  a  =  g  = 9,8 m/s^2






 y = 0
 y0 = H:
Relevant Notes:
 t(2H/3) is the time I set it takes for the rock to drop from heigh H to h. It's not a funtction, or anything like that.
 U(H/3) is the speed I set that the rock has when it arrives at heigh h.
H = 3h > h: (1) => h = 3h  4,9 * m/s^2 * t(2H/3)^2 <=> t(2H/3)^2 = 0,4 * s^2/m * h <=>
<=> t(2h/3) = 0,63 * s/sqrt(m) * sqrt(h)
(2) => U(2H/3) = 4,9 * m/s^ * t(2H/3)
y0 = h:
Relevant Notes:
 U0 in this case is U(2H/3.
t(h) is give to us by the book, which is 1,50 s.
h > 0: (1) => 0 = h + U(2H/3) * t(h) + 1/2 * a * t(h)^2 <=>
<=> 0 = h  4,9 * m/s^2 * t(2H/3) * 1,50 s  4,9 * m/s^2 * 2,25 * s^2 <=>
<=> h = 7,35 * m/s * t(2H/3) + 11,025 * m <=> h = 7,35 * m/s * [0,63 * s/sqrt(m) * sqrt(h)] + 11,025 * m
Now, I set sqrt(h) = u, therefore h = u^2
Also, I set sqrt(m) = n, so m = n^2
From where we left of:
u^2 = 4,63 * u * n^2/s * s/n + 11,025 * n^2 <=> u^2  4,63 * n * u 11,025 * n^2 = 0
Δ = β^2  4αγ = (4,63 * n)^2  4 * (1) * (11,025 * n^2) = 21, 43 * n62 + 44,1 * n^2 = 65,53 * n^2
u 1,2 = (+ 4,63 * n +/ sqrt(Δ))/2 = (4,63 +/ 8,09) * n/2 <=> u1 = 6,36 * n & u2 = 1,73 * n
Naturally, u1 is the correct answer, but u & n are variables I set myself. So:
sqrt(h) = 6,36 * sqrt(m)
I square both sides of the equation: h = 40,45 * m => H = 121,35 * m
So, if I did everything correctly, and din't miss anything or made a mistake in my thinking process, question (a) is done. What I need help with is question (b). Does anyone have any ideas what the second result could possibly be?
Thank you for your help and your time!
I searched around and found a manual, so this time I have the complete question in English (I'll use different numbers though, because mine is the 8th edition, and the manual is from the 14th, but it's not something that affects this exercise):
2.98) An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.50 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?
2. Homework Equations
y = y0 +U0*t + 1/2 * a * t^2 (1)
U = U0 + a * t (2)
3. The Attempt at a Solution
The assumption I made/parameters I took:
 ^ * Positive Direction is upwards.
  (+) * y = 0 is at the bottom of the cliff/the ground.
 * H = 3h
 * H is the whole height of the cliff.
 *  a  =  g  = 9,8 m/s^2






 y = 0
 y0 = H:
Relevant Notes:
 t(2H/3) is the time I set it takes for the rock to drop from heigh H to h. It's not a funtction, or anything like that.
 U(H/3) is the speed I set that the rock has when it arrives at heigh h.
H = 3h > h: (1) => h = 3h  4,9 * m/s^2 * t(2H/3)^2 <=> t(2H/3)^2 = 0,4 * s^2/m * h <=>
<=> t(2h/3) = 0,63 * s/sqrt(m) * sqrt(h)
(2) => U(2H/3) = 4,9 * m/s^ * t(2H/3)
y0 = h:
Relevant Notes:
 U0 in this case is U(2H/3.
t(h) is give to us by the book, which is 1,50 s.
h > 0: (1) => 0 = h + U(2H/3) * t(h) + 1/2 * a * t(h)^2 <=>
<=> 0 = h  4,9 * m/s^2 * t(2H/3) * 1,50 s  4,9 * m/s^2 * 2,25 * s^2 <=>
<=> h = 7,35 * m/s * t(2H/3) + 11,025 * m <=> h = 7,35 * m/s * [0,63 * s/sqrt(m) * sqrt(h)] + 11,025 * m
Now, I set sqrt(h) = u, therefore h = u^2
Also, I set sqrt(m) = n, so m = n^2
From where we left of:
u^2 = 4,63 * u * n^2/s * s/n + 11,025 * n^2 <=> u^2  4,63 * n * u 11,025 * n^2 = 0
Δ = β^2  4αγ = (4,63 * n)^2  4 * (1) * (11,025 * n^2) = 21, 43 * n62 + 44,1 * n^2 = 65,53 * n^2
u 1,2 = (+ 4,63 * n +/ sqrt(Δ))/2 = (4,63 +/ 8,09) * n/2 <=> u1 = 6,36 * n & u2 = 1,73 * n
Naturally, u1 is the correct answer, but u & n are variables I set myself. So:
sqrt(h) = 6,36 * sqrt(m)
I square both sides of the equation: h = 40,45 * m => H = 121,35 * m
So, if I did everything correctly, and din't miss anything or made a mistake in my thinking process, question (a) is done. What I need help with is question (b). Does anyone have any ideas what the second result could possibly be?
Thank you for your help and your time!