Simple(?) Kinematics Problem

[dontbeanegaton]

---

Members don't see this ad.

Simple question, but I don't get it:
A ball is shot horizontally from a rooftop. If you double the acceleration due to gravity, what happens to the distance the ball travels?
Kaplan's answer: The ball travels half the distance because the time it takes to hit the ground is cut in half.
How is this correct?
Vertical Distance = 1/2*a*t^2... If you double acceleration, you do not cut time in half. What am I missing? 😕

 

[Sports Junky]

Simple question, but I don't get it:
A ball is shot horizontally from a rooftop. If you double the acceleration due to gravity, what happens to the distance the ball travels?
Kaplan's answer: The ball travels half the distance because the time it takes to hit the ground is cut in half.
How is this correct?
Vertical Distance = 1/2*a*t^2... If you double acceleration, you do not cut time in half. What am I missing? 😕

You are right, I do not think the time simply is cut in half.

Derived equation of time in free-fall: t = (2h/g)^1/2

Therefore the time is reduced by a factor of (2^1/2)/2 or approximately 0.7

If the launch velocity remains the same, then the distance should be reduced to 0.7 of the original range.

Either there is some missing information based on the question stem, or the answer is simply incorrect. Where did you find the question anyways?

 

[LoLCareerGoals]

Let's see d_vert = 1/2*g*t^2, t = sqrt(2*d_vert/g), so t is 1/sqrt(2)ed like you guys discussed.

Now on to horizontal distance traveled: d_horz = 1/2*t_new^2*a0 = half of what it was before.
Maybe they meant t^2 is halfed? The final answer is correct imo, but the reasoning is a bit off.

 

[dontbeanegaton]

You are right, I do not think the time simply is cut in half.

Derived equation of time in free-fall: t = (2h/g)^1/2

Therefore the time is reduced by a factor of (2^1/2)/2 or approximately 0.7

If the launch velocity remains the same, then the distance should be reduced to 0.7 of the original range.

Either there is some missing information based on the question stem, or the answer is simply incorrect. Where did you find the question anyways?

It was a Kaplan FL problem in FL 6! I've since discovered a couple other questions that I think are just flat out wrong. There was no other information in the question stem.
I'm not thrilled with wasting time on poorly edited content, Kaplan.

 

[yahsha33]

Simple question, but I don't get it:
A ball is shot horizontally from a rooftop. If you double the acceleration due to gravity, what happens to the distance the ball travels?
Kaplan's answer: The ball travels half the distance because the time it takes to hit the ground is cut in half.
How is this correct?
Vertical Distance = 1/2*a*t^2... If you double acceleration, you do not cut time in half. What am I missing? 😕

Perhaps it is because acceleration due to gravity is not effective in the x direction? X direction deals with obejcts moving from the right to the left while acceleration deals with objects moving in the y-direction? In that case the time spent in the x-direction should not change. This may not be correct logic, but its just a thought.

 

[Jwinsler7]

Treat x and y components separately.

For the vertical component, g is doubled. y(vertical distance)=1/2at^2. Vertical y distance is constant, so only t will change.
Since g is x2, a^t is 1/2 the original value.

x=v1+1/2at^2
since t^2 is halved, x will be halved.

 

[Jwinsler7]

Perhaps it is because acceleration due to gravity is not effective in the x direction? X direction deals with obejcts moving from the right to the left while acceleration deals with objects moving in the y-direction? In that case the time spent in the x-direction should not change. This may not be correct logic, but its just a thought.

You're right about gravity not affecting the horizontal velocity. But time variable is the same for both x component and y component.

 

[LianaStan]

Treat x and y components separately.

For the vertical component, g is doubled. y(vertical distance)=1/2at^2. Vertical y distance is constant, so only t will change.
Since g is x2, a^t is 1/2 the original value.

x=v1+1/2at^2
since t^2 is halved, x will be halved.

The x distance is not affected by acceleration.

OP, can you specify which problem in Kap FL 6, I could not find it.

 

[Jwinsler7]

The x distance is not affected by acceleration.

OP, can you specify which problem in Kap FL 6, I could not find it.

It is affected by change in acceleration (gravity in this case).
Read my reasoning above.

 

[LianaStan]

The reasoning does not make sense because x=vt + 1/2at^2 is not the equation to use for horizontal motion. The horizontal component experiences no net force in the x direction, therefore there is no acceleration assuming negligible air resistance. X=vt is the correct equation to use. While gravity will affect the horizontal distance, this is through the indirect action of gravity affecting the time by which the object drops. Your equation heavily implies that you think the gravity will directly affect the horizontal distance traveled, which is not true.

 

[Jwinsler7]

The reasoning does not make sense because x=vt + 1/2at^2 is not the equation to use for horizontal motion. The horizontal component experiences no net force in the x direction, therefore there is no acceleration assuming negligible air resistance. X=vt is the correct equation to use. While gravity will affect the horizontal distance, this is through the indirect action of gravity affecting the time by which the object drops. Your equation heavily implies that you think the gravity will directly affect the horizontal distance traveled, which is not true.

You're right. I forgot ax would be 0. =/ For some reason, I thought of it as 1 for being constant, my fault.
OP, Kaplan probably used the same reasoning as I did for this question. But, the new horizontal distance should be 1/sqrt2 of the original distance rather than halved.

 

[LiFePO4try]

It was a Kaplan FL problem in FL 6! I've since discovered a couple other questions that I think are just flat out wrong. There was no other information in the question stem.
I'm not thrilled with wasting time on poorly edited content, Kaplan.

I agree that mistakes in nonfiction are a serious problem, as teaching books go through tons of editors, and unlike with fiction where people get frustrated with mistaken facts, nonfiction conveys facts to a novice (in the subject matter) most of the time and can’t contain misleading information. I get frustrated when I see, for example, wrong sucrose depictions in publications that I shouldn’t. . . . This is the stuff that we’re learning, and worse when it sticks with us (and we have to relearn)! In this case, however, I don’t think there was a mistake, and I’ll go through it now in this next post. [Yeay, it’s my first post!!!🙂]

 

[LiFePO4try]

Simple question, but I don't get it:
A ball is shot horizontally from a rooftop. If you double the acceleration due to gravity, what happens to the distance the ball travels?
Kaplan's answer: The ball travels half the distance because the time it takes to hit the ground is cut in half.
How is this correct?
Vertical Distance = 1/2*a*t^2... If you double acceleration, you do not cut time in half. What am I missing? 😕

So, to start with, I'll define things according to this rep.

and, like you said, H= 0+1/2*a*t^2 in air, where a=g.

The initial instinct is to double only the acceleration and see how H changes, but looking closer we see that t in air is dependant on acceleration, i.e it's a function of g by itself! Specifically, t in air= Vy / g, based on the 1st kinematic eqn.: Vy = Voy + a*t in air . [We could also derive this fact from the second eqn, as y= 1/2*g*t^2 => t = (2/g)(y/t) & Vavg = 1/2(Vy+0) Notice that y/t is Vavg] So, time in air is dependant on final velocity in the y direction and gravity. Put all together,
H= 1/2g (Vy/g)^2 => H=1/2(Vy^ 2 / g) we can now clearly see that doubling g will half the height.

P.S. sorry that the formatting is off on the equations.

 

[911 Turbo]

wrong

 

[milski]

Yes, the answer in Kaplan is wrong.
In the initial case, you have y=a*t^2/2 or t=sqrt(2y/a) and x=Vx*sqrt(2y/a)
For twice the acceleration, y=2a*t^2/2 or t=sqrt(y/a) and x=Vx*sqrt(y/a).

It's obvious that the new distance is sqrt(2) shorter.

Venom5, Vy is different in the two cases - you cannot compare the distances when expressed as function of it unless you first figure out what is the relation between the two Vy-s.

 

[911 Turbo]

Yes, the answer in Kaplan is wrong.
In the initial case, you have y=a*t^2/2 or t=sqrt(2y/a) and x=Vx*sqrt(2y/a)
For twice the acceleration, y=2a*t^2/2 or t=sqrt(y/a) and x=Vx*sqrt(y/a).

It's obvious that the new distance is sqrt(2) shorter.

Venom5, Vy is different in the two cases - you cannot compare the distances when expressed as function of it unless you first figure out what is the relation between the two Vy-s.

Ahh i see