Kinematics equation problem. (two stones dropped from different heights)

[Sentience]

Homework Statement

Two identical stones are dropped from rest and feel no air resistance as they fall. Stone A is dropped from height h, and stone B is dropped from height 2h. If stone A takes time t to reach the ground, stone B will take time:

a. 2t
b. t/2
c. 4t
d. t /(square root of t)
e. 2 * (Square root of t)

Homework Equations

V = Vo + at
X - Xo = Vo*t + .5at^2
V^2 = Vo^2 + 2a(X - Xo)

The Attempt at a Solution

The answer is e, 2 times square root of t. My teacher hinted that I use one of the kinematic equations and solve for 2h using 2t. However, because this is a generic problem, I'm not really sure how to go about it. It's got to be the 1st for the 3rd equation, I'm don't feel like I have enough information. Any advice/hints/answers would be appreciated.

Also, if I have other physics questions about different kinds of problems (say vectors or acceleration) should I just make one thread containing them all or separate threads?

 

[Modulated]

d. t /(square root of t)
e. 2 * (Square root of t)

Please check your problem sheet here - these two are not sensible options (as the square root of a time is not a time - how long is the square root of a second?). Are you sure it doesn't say something like (square root of 2) * t?

Homework Equations

V = Vo + at
X - Xo = Vo*t + .5at^2
V^2 = Vo^2 + 2a(X - Xo)

Well, these aren't all as relevant as each other. What do you know and what are you trying to find out? (In this context saying, e.g., the height is "h" counts as "knowing" it.)

Also, if I have other physics questions about different kinds of problems (say vectors or acceleration) should I just make one thread containing them all or separate threads?

Separate threads keep things simpler.

 

[Sentience]

Answers d and e should read:

d. t / (square root of 2)

e. t * (square root of 2)

I'm sorry for the confusion.

 

[Modulated]

Not at all - I was concerned you might have confused yourself! :-)

 

[Sentience]

Ok, looking at equation number 2, the change in position is h. Since it's from rest initial velocity * t cancels to zero, leaving me with h = 4.9m/s^2 * t^2. Here is where I'm stuck.

 

[Modulated]

That looks like a good start to me. So you know that h is proportional to t^2. If we now double h, we need to double the right-hand side of the equation too... so what would that mean we do to t?

 

[Sentience]

When I solve for t, I have h divided by 4.9m/s^2 all under a radical.

If I made h 2h instead, on the time side I would have to multiply t by root 2 to obey the rules of algebra.

Is this good reasoning?

 

[Modulated]

Absolutely - well done.

 

[Sentience]

Thanks for the help modulated.