# Kinematics questions?

## Homework Statement

1. Same objects, object 1 takes t seconds to hit the ground, and object 2 takes 2t seconds. If object 1 was dropped at h meters, where was object 2 dropped?

2. Same objects, object 1 was dropped at h meters, object 2 was dropped at 2h meters. if object 1 takes t seconds to hit the ground. how long will it take for object 2 to hit the ground?

3. A person throw a rock on earth, and it returns to his hand after t seconds. Then he went to planet x and did the exact same thing, but this time, the rock takes 2t seconds. What's the acceleration due to gravity on planet x? In terms of g.

Disregard air resistance in all three questions

## The Attempt at a Solution

cepheid
Staff Emeritus
Gold Member
Welcome to PF Vhsmith37!

## Homework Statement

1. Same objects, object 1 takes t seconds to hit the ground, and object 2 takes 2t seconds. If object 1 was dropped at h meters, where was object 2 dropped?

2. Same objects, object 1 was dropped at h meters, object 2 was dropped at 2h meters. if object 1 takes t seconds to hit the ground. how long will it take for object 2 to hit the ground?

3. A person throw a rock on earth, and it returns to his hand after t seconds. Then he went to planet x and did the exact same thing, but this time, the rock takes 2t seconds. What's the acceleration due to gravity on planet x? In terms of g.

Disregard air resistance in all three questions

## The Attempt at a Solution

All three of these questions can be answered if you know the answer to this question that follows: what is the expression for distance vs. time for an object moving with constant acceleration?

(heres a nice equation)

$$\Delta z=-\frac{1}{2}gt^2$$

You can use this when an object has a force mg in the negative z direction, and when it starts from rest with only gravity acting upon it (e.g. free fall)

in general, if an object starts with a velocity V0 in the z direction and starts at position Z0 then we have the equation:

$$z=z_0 + V_0 t - \frac{1}{2}gt^2$$

which gives the position z at any time t

cepheid
Staff Emeritus
Gold Member
(heres a nice equation)

$$\Delta z=-\frac{1}{2}gt^2$$

You can use this when an object has a force mg in the negative z direction, and when it starts from rest with only gravity acting upon it (e.g. free fall)

in general, if an object starts with a velocity V0 in the z direction and starts at position Z0 then we have the equation:

$$z=z_0 + V_0 t - \frac{1}{2}gt^2$$

which gives the position z at any time t

The point was sort of to have the OP come up with this answer him/herself. :grumpy:

Thank you all for answering me question, but I'm still confused... sorry, can you please explain further? i know one of them is 4 times the other one, just dont remember which, and why 4X

cepheid
Staff Emeritus
Gold Member
Thank you all for answering me question, but I'm still confused... sorry, can you please explain further? i know one of them is 4 times the other one, just dont remember which, and why 4X

You can use the first equation that AlexChandler gave you to understand all of this. For the first question you were asked:

If time interval is equal to t, then what is Δz equal to? (This is an obvious question, since the equation tells you). Now, what happens if you substitute in 2t instead of just t? How does the new Δz compare to the old Δz? By what factor is it larger?

For the second question you were asked:

If the displacement, Δz, is equal to h, then when you plug this in, what is t equal to (according to the equation)?

Now if you increase the displacement to 2h, what is t now equal to? By what factor has it changed from before? (To answer this, compare your two expressions for t that you obtained in the two cases of Δz = h and Δz = 2h. How much larger is the second t than the first t?)

For the third question you were asked, if the height of the throw remains the same, but somehow the time interval is doubled, by how what factor must g change (in the equation) in order to make this happen?