Simple Projectile Motion of a rock

[cowmoo32]

Homework Statement

A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

Homework Equations

v = v⁰ + at²
h = v₀t + 0.5at²
v² = v₀² + 2ahh = distance in this case

The Attempt at a Solution

t for $\frac{h}{3}$ = 1 sec

$\frac{h}{3}$ = v₀(1) + 0.5(-9.81)(1²)
.
.
.
h = 3v_o + 14.715$\frac{m}{sec^{2}}$

I've re-arranged the equations but I'm always left with 2 unknowns. I understand this is basic stuff, so just a bump in the right direction will be much appreciated.

 

[hotvette]

I believe you've not interpreted the problem statement correctly. I think it says the last 1/3 of the distance is covered in 1 second, not the first 1/3. If you think about t₁ as the time at which the object has covered 2/3 of the distance and t₂ as the time it has covered the entire distance, you can write two equations use the fact that t₂ - t₁ = 1 to solve for h.

Also, what do you know about v₀ from the original problem statement?

 

[NoPoke]

Try making a drawing and identifying all the information that you know exactly and all the information that is dependent upon something else. Pick a consistent set of subscripts that makes use of the information that you know.

Do you know the time that the object hits the ground ? Do you know the time when it has the bottom 1/3 of the cliff remaining?

 

[azizlwl]

Homework Statement

A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

Homework Equations

v = v⁰ + at²
h = v₀t + 0.5at²
v² = v₀² + 2ah


h = distance in this case

The Attempt at a Solution

t for $\frac{h}{3}$ = 1 sec

$\frac{h}{3}$ = v₀(1) + 0.5(-9.81)(1²)
.
.
.
h = 3v_o + 14.715$\frac{m}{sec^{2}}$

I've re-arranged the equations but I'm always left with 2 unknowns. I understand this is basic stuff, so just a bump in the right direction will be much appreciated.

You should know the intial velocity and total distance covered.
And also distance covered 2/3 of h in less than 1 sec of total time.

 

[cowmoo32]

I believe you've not interpreted the problem statement correctly. I think it says the last 1/3 of the distance is covered in 1 second, not the first 1/3. If you think about t₁ as the time at which the object has covered 2/3 of the distance and t₂ as the time it has covered the entire distance, you can write two equations use the fact that t₂ - t₁ = 1 to solve for h.

Also, what do you know about v₀ from the original problem statement?

v₀=0 for total h. But v₀ for h/3 is v_{t after 2h/3} And it does cover the last third in one second.

A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall.

You should know the intial velocity and total distance covered.
And also distance covered 2/3 of h in less than 1 sec of total time.

How do you know that?

 

[cowmoo32]

Another attempt. Is this correct? I feel like the math is ok.

V_f = V_i² + 2ah
V_f = 0 + 2(9.81)(h)
V_f = 4.43h

I'm calling V at $\frac{2h}{3}$ V₂

V₂ = 4.43$\frac{2h}{3}$
V₂ = 2.953h

Using the equation from my first post but V₂ instead of V₀:

h = 3V₂ - 14.715
Substitute V₂ = 2.953h
h = 28.59h - 14.715
h = 1.875 meters

 

[CAF123]

I got an answer of 1.4865m doing the problem 2 different ways. This yielded a quadratic in h which gave me this answer and ≈145m both times. I am still trying to figure out the physical significance of these 2 results. I feel 1.4865m is a bit small for a cliff.

 

[johnqwertyful]

Another attempt. Is this correct? I feel like the math is ok.

V_f = V_i² + 2ah
V_f = 0 + 2(9.81)(h)
V_f = 4.43h

I'm calling V at $\frac{2h}{3}$ V₂

V₂ = 4.43$\frac{2h}{3}$
V₂ = 2.953h

Using the equation from my first post but V₂ instead of V₀:

h = 3V₂ - 14.715
Substitute V₂ = 2.953h
h = 28.59h - 14.715
h = 1.875 meters

I think you missed a square in your formula. Also how you got from 2(9.81) to 4.43 I don't know.

 

[cowmoo32]

I think you missed a square in your formula. Also how you got from 2(9.81) to 4.43 I don't know.

I forgot to square V_f in the formula when I typed it out; I have it correct on my paper. (2*9.81)^.5 = 4.43

 

[hotvette]

There is an easy sanity check on your answer. From h you can calculate the total time it takes for the object to fall. Fyi, my result was h > 100m.

 

[johnqwertyful]

I forgot to square V_f in the formula when I typed it out; I have it correct on my paper. (2*9.81)^.5 = 4.43

Then you need a sqrt(h)

 

[NoPoke]

Still looks to me like you are getting confused over what all the letters in the algebra are. To solve this you need to be quite careful with your various vs, hs, ts.

To get a ball park idea of the cliff height you are seeking why not just fill in a table of where the rock is as it descends the cliff? using g = 10m/sec/sec

t fallen
1 5m
2 20m
3 45m
4 80m
5 125m
6 180m

So for any given time you have the height and in the previous row the height it would have been at one second before. None of those are exactly 2/3rds but one is close and gives you an easy check on your algebra.

 

[CAF123]

. Fyi, my result was h > 100m.

Did you also get another solution by solving a quadratic? As I said in my last post, I got ≈1.5m and ≈145m. What is the significance of me getting the 1.5m solution?

 

[azizlwl]

h=0.5at²
(2/3)h=0.5a(t-1)²

t=5.5sec and 0.5sec.

I guess only for t >1sec is applicable.

 

[CAF123]

h=0.5at2
(2/3)h=0.5a(t-1)2

t=5.5sec and 0.5sec.

I guess only for t >1sec is applicable.

Are you answering me? The 1.5m solution might have been the result of a square root fallacy, generating incorrect solutions because I had to square at one point. Also, I inputted these two solutions into another eqn I had formed, and the 145m soln satisfied this while the lower soln did not.