Physics gravity problem

[pabloyikes]

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A ball is shot horizontally off a cliff. What would happen to the horizontal distance it travels if the acceleration due to gravity were doubled and all other factors remain the same?


I got a problem similar to this from Kaplan and it claimed the answer is that the distance would be halved. Wouldn't the distance be decreased by a factor of sqrt2 ?

I set:
d(vertical) = d(vertical)2 [i.e. height of cliff]

1/2at^2 = 1/2 (2a) t'^2

t^2/2 = t'^2

t' = t/sqrt2

d(horizontal) = N * d(horizontal)2

vt = N v(t/sqrt2)

N = sqrt2


Please help me understand how/if I'm wrong.

Thanks in advance,
P

 

[Chrisz]

Y=0.5gt^2
If g doubled, with the assumption of same initial Vx
Y=gt*^2

(t*/t)^2=0.5

t*/t=sqroot(0.5)

So if g is doubled, t* is indeed smaller by a factor of sqroot(0.5), and horizontal distance will be exactly shorten by the same factor if initial horizontal velocity is the same before and after the change.

If you did not miss any further information in the passage or question stem, I would say your explanation is correct. It might have a typo, or the author write up the answer too fast.

 

[DrknoSDN]

I don't believe your wrong in solving the posed question. I solved a different way and got the same answer. Perhaps it's your interpretation of the question? I only suggest that because you said you got a "similar" problem from kaplan, so maybe variation in wording changed the problem.

Anyway, my solution:
Horizontal distance is proportional to horizontal velocity * time.

Solve for time with respect to height:
X=Xo + 1/2 at^2
t=root(2x/a)

Double (a) results in a change of factor of root(0.5)

 

[blackmi4]

Forget equations for a moment.

You split up velocity in the x direction and the y direction.

Distance in the x direction depends solely on time in the air.

Time in the air is determined by the initial vertical velocity and the force of gravity acting downwards against the object.

If you double the amount of force acting downwards against the vertical velocity, the vertical velocity changes twice as fast.

If vertical velocity changes (decreases) twice as fast, then you intuitively know that time in the air will be cut in half.

1/2 the time in the air results in 1/2 the horizontal distance traveled.

 

[Chrisz]

Forget equations for a moment.

You split up velocity in the x direction and the y direction.

Distance in the x direction depends solely on time in the air.

Time in the air is determined by the initial vertical velocity and the force of gravity acting downwards against the object.

If you double the amount of force acting downwards against the vertical velocity, the vertical velocity changes twice as fast.

If vertical velocity changes (decreases) twice as fast, then you intuitively know that time in the air will be cut in half.

1/2 the time in the air results in 1/2 the horizontal distance traveled.

No offense, if you was correct, your explanation could have been logically demonstrated by formulas, unfortunately you are making a logical mistake by using ur intuitive reasoning. Your assumption, if you double the gravity, velocity at each corresponding height would be doubled as well. I am pointing out ur mistake right here. The vertical velocity is doubled at a corresponding "TIME", not a corresponding "HEIGHT". Now you see the difference.

 

[DrknoSDN]

If vertical velocity changes (decreases) twice as fast, then you intuitively know that time in the air will be cut in half.

No.

I scoured the internet looking for a calculator where you can plug and chug examples if you wanted. (set air resistance to zero)
http://keisan.casio.com/exec/system/1231475371

 

[blackmi4]

No offense, if you was correct, your explanation could have been logically demonstrated by formulas, unfortunately you are making a logical mistake by using ur intuitive reasoning. Your assumption, if you double the gravity, velocity at each corresponding height would be doubled as well. I am pointing out ur mistake right here. The vertical velocity is doubled at a corresponding "TIME", not a corresponding "HEIGHT". Now you see the difference.

No.

I scoured the internet looking for a calculator where you can plug and chug examples if you wanted. (set air resistance to zero)
http://keisan.casio.com/exec/system/1231475371

Can we agree that the time in the air determines the distance traveled in the horizontal direction?

Lets pretend the ball is shot upward at 5 meters per second. It will also hit the ground at 5 meters per second.

That is vyi=5 m/s, vyf=-5 m/s.

dv=-10m/s

dv=a*dt
-10m/s=-10m/s^2*dt
1s=dt

If gravity is doubled:

dv=a*dt
-10m/s=-20m/s^2*t
.5=dt

 

[Chrisz]

Can we agree that the time in the air determines the distance traveled in the horizontal direction?

Lets pretend the ball is shot upward at 5 meters per second. It will also hit the ground at 5 meters per second.

That is vyi=5 m/s, vyf=-5 m/s.

dv=-10m/s

dv=a*dt
-10m/s=-10m/s^2*dt
1s=dt

If gravity is doubled:

dv=a*dt
-10m/s=-20m/s^2*t
.5=dt

Hi, bro, I intentionally capitalized TIME and HEIGHT in my original statement. The original poster defines the maximum height to be the cliff, which does not change regardless of whether you double gravity or not. In your example, When you shoot the ball upward, the maximum height will be lowered after the gravity is doubled. You truly understand what you talking about. If you still have difficulty understand the question, simply derive a formula while holding the maximum height constant and doubling the gravity. Intuitive reasoning is good to only limited extent, and you have to into account all the conditions.

 

[blackmi4]

Hi, bro, I intentionally capitalized TIME and HEIGHT in my original statement. The original poster defines the maximum height to be the cliff, which does not change regardless of whether you double gravity or not. In your example, When you shoot the ball upward, the maximum height will be lowered after the gravity is doubled. You truly understand what you talking about. If you still have difficulty understand the question, simply derive a formula while holding the maximum height constant and doubling the gravity. Intuitive reasoning is good to only limited extent, and you have to into account all the conditions.

To be fair I was kind of lazy here in that I actually didn't read the OP's full message. All I saw was projectile motion problem with everything the same except doubling gravity and him/her asking why distance cuts in half.

I'll explain his/her problem specifically though -- the ball shooting horizontally off a cliff -- with plain intuition. Plain intuition works in both cases.

The ball is shot, horizontally, off a cliff. The final distance in the y direction doesn't change. Initial vertical velocity is 0.

What forces act in the y direction? Just gravity. There is no air resistance/drag. Since air resistance/drag are what cause terminal velocity, and they are absent, the object just goes faster and faster.

What is acceleration? The change in velocity over the change in time.

Again, if we double acceleration then the velocity changes twice as fast. That means that average velocity has doubled. That means that the object covers twice the distance in the same amount of time.

If twice the distance is covered in the same amount of time, and the distance from the cliff to the ground doesn't change (it doesn't) then the object will reach the ground in half the time.

If the object reaches the ground in half the time then the horizontal distance will also cut in half.

Do you really need equations to understand that? I don't think so. But we can use one simple one.

x=.5gt^2.

As we all know, in one second an object (acted on in the y direction by just the force of gravity) will travel 5 meters in the vertical direction.

x=.5*10*1^2. x=5m

Two seconds? 20 meters.

x=.5*10*2^2.=20m

Now what happens if we double gravity?

x=.5*20*1^2
x=10m in the vertical direction after 1 second.

x=.5*20*2^2
x=40m in the vertical direction after 2 seconds.


So, now you can see that twice the distance is covered in the same amount of time. So you know that the object will cover the distance between the cliff and the ground in 1/2 the time.

Again, we don't need to use equations here if we just use our logic and intuition.

 

[Chrisz]

To be fair I was kind of lazy here in that I actually didn't read the OP's full message. All I saw was projectile motion problem with everything the same except doubling gravity and him/her asking why distance cuts in half.

I'll explain his/her problem specifically though -- the ball shooting horizontally off a cliff -- with plain intuition. Plain intuition works in both cases.

The ball is shot, horizontally, off a cliff. The final distance in the y direction doesn't change. Initial vertical velocity is 0.

What forces act in the y direction? Just gravity. There is no air resistance/drag. Since air resistance/drag are what cause terminal velocity, and they are absent, the object just goes faster and faster.

What is acceleration? The change in velocity over the change in time.

Again, if we double acceleration then the velocity changes twice as fast. That means that average velocity has doubled. That means that the object covers twice the distance in the same amount of time.

If twice the distance is covered in the same amount of time, and the distance from the cliff to the ground doesn't change (it doesn't) then the object will reach the ground in half the time.

If the object reaches the ground in half the time then the horizontal distance will also cut in half.

Do you really need equations to understand that? I don't think so. But we can use one simple one.

x=.5gt^2.

As we all know, in one second an object (acted on in the y direction by just the force of gravity) will travel 5 meters in the vertical direction.

x=.5*10*1^2. x=5m

Two seconds? 20 meters.

x=.5*10*2^2.=20m

Now what happens if we double gravity?

x=.5*20*1^2
x=10m in the vertical direction after 1 second.

x=.5*20*2^2
x=40m in the vertical direction after 2 seconds.


So, now you can see that twice the distance is covered in the same amount of time. So you know that the object will cover the distance between the cliff and the ground in 1/2 the time.

Again, we don't need to use equations here if we just use our logic and intuition.

Let me show you the proper calculation for it. We define the maximum height as Y, which does not change before and after the cliff.
Y=0.5gt^2
if you double gravity
Y=0.5(2g)t*^2=gt*^2
--->
t=sqroot(2Y/g)
t*=sqroot(Y/g)
--->
t*/t=sqroot(0.5)

now you tell me if t* is 0.5 of t, or sqroot(0.5) of t.

If you dont trust me, it is fine. Go ask your professor. I spent too much time on this question already.

 

[blackmi4]

Let me show you the proper calculation for it. We define the maximum height as Y, which does not change before and after the cliff.
Y=0.5gt^2
if you double gravity
Y=0.5(2g)t*^2=gt*^2
--->
t=sqroot(2Y/g)
t*=sqroot(Y/g)
--->
t*/t=sqroot(0.5)

now you tell me if t* is 0.5 of t, or sqroot(0.5) of t.

If you dont trust me, it is fine. Go ask your professor. I spent too much time on this question already.

I am beginning to see your point. If you set up the equation as you said then time isn't doubled or halved or whatever I said. But that still doesn't explain the answer.

If time in the air determines horizontal distance, how can you double gravity and double horizontal distance?

You have me doubting myself now but again, assuming the answer in the book is correct, how can someone explain it?

If the book is incorrect then the writer could hand been making my mistake in regards to time in the air.

 

[Chrisz]

I am beginning to see your point. If you set up the equation as you said then time isn't doubled or halved or whatever I said. But that still doesn't explain the answer.

If time in the air determines horizontal distance, how can you double gravity and double horizontal distance?

You have me doubting myself now but again, assuming the answer in the book is correct, how can someone explain it?

If the book is incorrect then the writer could hand been making my mistake in regards to time in the air.

Let me give you little bit more detail just to make it clear.the horizontal distance is not doubled. In this case, horizontal distance is shortened by a factor of sqroot(0.5). Because when the gravity is doubled, the falling object reaches the ground faster with time shortened by sqroot(0.5). And horizontal distance is dependent on initial horizontal velocity and time traveled. x=sqroot(0.5)tVx.

One more thing to make you better understand the concept.

Think of this statement. Y is linearly proportional to t^2, not linearly proportional to t.

Since you like numerical examples, which I personally not fond of, I give you one to let you better appreciate the above statement

You say, if you double g, time spent would be halved. If this statement is true, an object will fall the same vertical distance in half amount of time

Let's say t=1 with g=9.8
Y=0.5(9.8)1^2=4.9

Now you set t=0.5 with g=2(9.8)
Y=9.8(0.5)^2=2.45

You see with half amount of time while having g doubled, it does not travel the same distance.

But if you set t=sqroot(0.5)

Y=9.8[sqroot(0.5)]^2=4.9 it becomes 4.9 again.

A squre function is a parabola. It increases slowly before t=1, and increase becomes substantial after t=1

Books have typos all the time, that is why you see some books with new editions coming out. I use Berkeley Review 2010 ed, I found few conceptual mistakes in it. I believe largely due to not being careful enough

 

[dudewheresmymd]

A ball is shot horizontally off a cliff. What would happen to the horizontal distance it travels if the acceleration due to gravity were doubled and all other factors remain the same?


I got a problem similar to this from Kaplan and it claimed the answer is that the distance would be halved. Wouldn't the distance be decreased by a factor of sqrt2 ?

I set:
d(vertical) = d(vertical)2 [i.e. height of cliff]

1/2at^2 = 1/2 (2a) t'^2

t^2/2 = t'^2

t' = t/sqrt2

d(horizontal) = N * d(horizontal)2

vt = N v(t/sqrt2)

N = sqrt2


Please help me understand how/if I'm wrong.

Thanks in advance,
P

If this were a traditional kinematics problem with an angle and some component of vy, the answer would be halved as Kaplan alludes to. However, b/c the velocity is purely horizontal, we use

y=1/2at^2
sqrt(2y/g)=t
Doubling g would decrease t by a factor of sqrt(2) therefore the horizontal distance (vx * time in air) would decrease by a factor of sqrt(2) as vx is constant and t decreases by sqrt(2).

 

[Chrisz]

T


If this were a traditional kinematics problem with an angle and some component of vy, the answer would be halved as Kaplan alludes to. However, b/c the velocity is purely horizontal, we use

y=1/2at^2
sqrt(2y/g)=t
Doubling g would decrease t by a factor of sqrt(2) therefore the horizontal distance (vx * time in air) would decrease by a factor of sqrt(2) as vx is constant and t decreases by sqrt(2).

I dont think this works in every case. It only works if the initial vertical velocity is held constant, so i would not call it call traditional, and the maximum height would be lowered as well. It is better to solve physics questions on conditional basis

 

[dudewheresmymd]

I dont think this works in every case. It only works if the initial vertical velocity is held constant, so i would not call it call traditional, and the maximum height would be lowered as well. It is better to solve physics questions on conditional basis

Ah yes, traditional is poor word choice. I meant in other kinematics questions where we are given an angle and an initial velocity that is held constant.. good catch.

 

[syoung]

To use numbers...
At the same height on two different planets A and B with g(A) = 10 m/s^2 and g(B) = 20 m/s^2, with zero air resistance and zero vertical initial velocity

Let 500 m be the height of the cliff
On Planet A
500 m = 1/2 (10m/s^2) * t^2
t = 10 s

On Planet B
500 m = 1/2 (20m/s^2) * t^2
t = 7.07 s

With same initial horizontal initial velocity of 10 m/s...

On Planet A
R = 10 m/s * 10 s
R = 100 m

On Planet B
R = 10 m/s * 7.07 s
R = 70.7 m

As you can see, the horizontal distance does not change by 1/2 if the gravity is doubled. Conceptually you can see this as well if you don't do numbers and wish to have a more mathematical approach. I tend to like plugging in an easy number and comparing right away. Solves any of the "intuition" dilemmas.

Also the Kaplan author may have had the original problem say 4x gravity instead of 2x gravity, which would lead to the answer of half the horizontal distance. They probably were lazy.

 

[blackmi4]

Let me give you little bit more detail just to make it clear.the horizontal distance is not doubled. In this case, horizontal distance is shortened by a factor of sqroot(0.5). Because when the gravity is doubled, the falling object reaches the ground faster with time shortened by sqroot(0.5). And horizontal distance is dependent on initial horizontal velocity and time traveled. x=sqroot(0.5)tVx.

One more thing to make you better understand the concept.

Think of this statement. Y is linearly proportional to t^2, not linearly proportional to t.

Since you like numerical examples, which I personally not fond of, I give you one to let you better appreciate the above statement

You say, if you double g, time spent would be halved. If this statement is true, an object will fall the same vertical distance in half amount of time

Let's say t=1 with g=9.8
Y=0.5(9.8)1^2=4.9

Now you set t=0.5 with g=2(9.8)
Y=9.8(0.5)^2=2.45

You see with half amount of time while having g doubled, it does not travel the same distance.

But if you set t=sqroot(0.5)

Y=9.8[sqroot(0.5)]^2=4.9 it becomes 4.9 again.

A squre function is a parabola. It increases slowly before t=1, and increase becomes substantial after t=1

Books have typos all the time, that is why you see some books with new editions coming out. I use Berkeley Review 2010 ed, I found few conceptual mistakes in it. I believe largely due to not being careful enough

You are right on here, chrisz. I am ashamed to say that my confidence on this problem was misplaced; I had this type of response coming to me. Thank you for spelling this out.

 

[September24]

I know we are done with this, but BR gives a formula:

R=((v^2)*(sin20))/g

Dont know how much this helps, but doesnt doubling gravity half the range?

 

[Chrisz]

You are right on here, chrisz. I am ashamed to say that my confidence on this problem was misplaced; I had this type of response coming to me. Thank you for spelling this out.

It is ok. No shame for that. Being confident is good. U are way better than some of the students when I was in college. They always refer to authority without having a reasonable explanation. I mean you are the type of person that, given a reasonable explanation,would more than happy to accept it. Once I was taking gen chem class, the professor made a conceptual mistake on a complicated gas pressure question on the exams most of the students got it wrong. When I made the explanation to them, they did not accept it because they argued the professor said so in class. Actually, my prof made a quick and intuitive reasoning when he solved the question. Don't feel bad. At least, you learned the concept. This is the most important thing

 

[Chrisz]

I know we are done with this, but BR gives a formula:

R=((v^2)*(sin20))/g

Dont know how much this helps, but doesnt doubling gravity half the range?

This equation is derived from the fact that initial velocity is assumed to be given. With a certain initial velocity with specific magnitude and angle, it can only reach certain amount of height. If you double gravity with initial velocity held constant, the range would be shortened by half. But how about the maximum height, the maximum height is lowered. But in the case of cliff, the maximum height is fixed. In order to keep the maximum height constant, you have to increase initial velocity(assuming going from the ground to cliff), which can be calculated backward from the cliff height. 2mgh+ 0.5m(vcos(theta))^2=0.5mv^2.
Or you can use 0.5(2)gt^2=h to find t. Then calculate the velocity 2gt=v Then, v/sin(theta) is the initial velocity. There still other ways to do this question. The point is max height does not change.