Vertical displacement with time of a projectile

[hello478]

Homework Statement

image below

Relevant Equations

suvat equations

my answer was D, which is correct considering the motion downwards eg, object thrown from a cliff
but what if an object is moving up?like a football being kicked?
i cant understand how this graph applies to that...

 

[kuruman]

What can you say about the components of the velocity in each direction from the 4 graphs?

 

[PeroK]

i cant understand how this graph applies to that...

It doesn't necessarily have to. It didn't say that a single graph must apply to all possible scenarios. If you were shown four pictures and asked which one was a dog, then you could identify the dog. But, it doesn't mean all dogs look like that. If it's a chihuahua, then it doesn't look like a St Bernard.

 

[PeroK]

my answer was D, which is correct considering the motion downwards eg, object thrown from a cliff

I would have said D is a rocket accelerating upwards.

 

[kuruman]

To what @PeroK said, I would add that none of the 4 plots shows the correct horizontal displacement vs. time. Do you see why that is the case?

 

[hello478]

To what @PeroK said, I would add that none of the 4 plots shows the correct horizontal displacement vs. time. Do you see why that is the case?

the correct horizontal displacement with time would be a straight line passing from origin

 

[kuruman]

the correct horizontal displacement with time would be a straight line passing from origin

RIght. Do you see any straight lines?

 

[hello478]

RIght. Do you see any straight lines?

no
but i was asking

but what if an object is moving up?like a football being kicked?
i cant understand how this graph applies to that...

i do know the answer i D
but can you explain how the graph relates to an object moving up in a projectile...

 

[kuruman]

but i was asking

I know what you were asking. I wanted to make sure that you understand why A and C cannot be correct because they lines are not straight.

Now how do you know that the answer is D? If you toss a rock with some initial horizontal and vertical velocity it will go across at constant velocity while, in the vertical direction, it goes up and then back down. Which of the two graphs looks like that is the case, B or D?

 

[haruspex]

my answer was D, which is correct considering the motion downwards eg, object thrown from a cliff

Are you saying that graphs B and D are inverted, i.e. that a downward displacement is positive?
How do you know that? It does not say so. If you choose instead that up is positive then that makes the correct answer B.

 

[PeroK]

Are you saying that graphs B and D are inverted, i.e. that a downward displacement is positive?
How do you know that? It does not say so. If you choose instead that up is positive then that makes the correct answer B.

Except that the question itself does not say "projectile motion". And, graph B has non-zero initial velocity.

 

[haruspex]

Except that the question itself does not say "projectile motion". And, graph B has non-zero initial velocity.

True, which means that when it says it moves from rest with uniform velocity horizontally it intends that that velocity is purely horizontal. I had read it as unspecified vertical velocity. With my reading and a vertical acceleration that could be up or down both B and D could be correct.

 

[PeroK]

True, which means that when it says it moves from rest with uniform velocity horizontally it intends that that velocity is purely horizontal. I had read it as unspecified vertical velocity. With my reading and a vertical acceleration that could be up or down both B and D could be correct.

I missed that. Another mess of a question.

 

[Mister T]

There is nothing wrong with the question. Since it moves from rest the initial slope of the graph must be zero. Choice B lacks that feature. It shows an object with a nonzero initial velocity.

 

[haruspex]

Since it moves from rest the initial slope of the graph must be zero.

It is not clear what is meant by "moves from rest". That conflicts with "constant horizontal velocity" unless that is zero.

 

[pbuk]

Calm down everyone, there is nothing wrong with this question.

• At time t = 0 the object is at rest.
• At time t > 0 the horizontal component of velocity is constant so the magnitude of the horizontal displacement increases linearly wrt time. This rules out answers A and C.
• At time t > 0 the magnitude of the vertical component of velocity is increasing at a constant rate so the magnitude of the displacement increases quadratically wrt time. This rules out answer B.
• This leaves the answer D, which can be interpreted either as the vertical displacement downwards of an object falling under gravity or as the vertical displacement upwards of a rocket accelerating upwards at a constant rate.

Note that answer B does meet the criterion "uniform acceleration vertically" where that acceleration opposes an initial velocity (e.g. for an object thrown upwards) but the question states that the object has no initial velocity: it is at rest.

 

[pbuk]

It is not clear what is meant by "moves from rest".

It means that at t = 0 velocity is zero.

That conflicts with "constant horizontal velocity"

No it doesn't, because this describes the situation when t > 0. Of course if the constant horizontal velocity is non-zero for t > 0 then horizontal acceleration at t = 0 would be infinite, but there is nothing in the question to rule this out, whereas if you want to interpret answer B as correct then vertical acceleration at t = 0 would be infinite, contradicting the stipulation of the question that vertical acceleration is constant.

 

[PeroK]

At time t = 0 the object is at rest.
At time t > 0 the horizontal component of velocity is constant zero so the magnitude of the horizontal displacement increases linearly wrt time is zero. This rules out answers A and C.

 

[haruspex]

At time t > 0 the magnitude of the vertical component of velocity is increasing

Not if there is an initial vertical velocity with the opposite sign to the acceleration.

 

[pbuk]

Not if there is an initial vertical velocity with the opposite sign to the acceleration.

The question states that there is no initial vertical velocity ("an object moves from rest").

 

[pbuk]

No, the horizontal component of velocity is not necessarily zero, see #17.

 

[pbuk]

In life we have to answer the questions we are given, not the questions that we wish we had been given.

As advisors on this forum we should remember this.

 

[PeroK]

It means that at t = 0 velocity is zero.

No it doesn't, because this describes the situation when t > 0. Of course if the constant horizontal velocity is non-zero for t > 0 then horizontal acceleration at t = 0 would be infinite, but there is nothing in the question to rule this out, whereas if you want to interpret answer B as correct then vertical acceleration at t = 0 would be infinite, contradicting the stipulation of the question that vertical acceleration is constant.

If you can have a non-zero initial horizontal component, then you can have a non-zero initial vertical component, in which case both B and D are valid.

The question is a mess.

 

[PeroK]

In life we have to answer the questions we are given, not the questions that we wish we had been given.

The students are entitled to better. It's not enough to say "I wrote the question and I understand what I meant and anyone who can't figure out my arbitrary carelessness with kinematic data gets no marks".

 

[pbuk]

If you can have a non-zero initial horizontal component, then you can have a non-zero initial vertical component, in which case both B and D are valid.

But you can't have a non-zero initial horizontal component ("an object moves from rest"), nor a non-zero initial vertical component, and so B is not valid.

 

[PeroK]

But you can't have a non-zero initial horizontal component ("an object moves from rest"), nor a non-zero initial vertical component, and so B is not valid.

That's what we all thought, until you starting telling us we were wrong:

Calm down everyone, there is nothing wrong with this question.

• At time t = 0 the object is at rest.
• At time t > 0 the horizontal component of velocity is constant so the magnitude of the horizontal displacement increases linearly wrt time. This rules out answers A and C.
• At time t > 0 the magnitude of the vertical component of velocity is increasing at a constant rate so the magnitude of the displacement increases quadratically wrt time. This rules out answer B.
• This leaves the answer D, which can be interpreted either as the vertical displacement downwards of an object falling under gravity or as the vertical displacement upwards of a rocket accelerating upwards at a constant rate.

Note that answer B does meet the criterion "uniform acceleration vertically" where that acceleration opposes an initial velocity (e.g. for an object thrown upwards) but the question states that the object has no initial velocity: it is at rest.

Your position is self-contradictory, as shown by the two contradictory statements I've underlined.

 

[pbuk]

Your position is self-contradictory, as shown by the two contradictory statements I've underlined.

1. At t = 0 the object is at rest
2. At t > 0 the object has constant horizontal velocity.

Where is the contradiction?

 

[PeroK]

1. At t = 0 the object is at rest
2. At t > 0 the object has constant horizontal velocity.

Where is the contradiction?

That contradicts your second assertion that the object has no vertical velocity for ##t > 0##.

 

[PeroK]

1. At t = 0 the object is at rest
2. At t > 0 the object has constant horizontal velocity.

Where is the contradiction?

And, with all due respect, I highlighted two statements in your post. Not just one.

 

[jbriggs444]

Calm down everyone, there is nothing wrong with this question.

• At time t = 0 the object is at rest.
• At time t > 0 the horizontal component of velocity is constant so the magnitude of the horizontal displacement increases linearly wrt time. This rules out answers A and C.
• At time t > 0 the magnitude of the vertical component of velocity is increasing at a constant rate so the magnitude of the displacement increases quadratically wrt time. This rules out answer B.

You cannot have it both ways. If being at rest at ##t=0## with a discontinuous change at ##t>0## is allowed for the horizontal velocity then surely the same can be true for the vertical velocity.

This makes the stipulation that the object is at rest at ##t=0## irrelevant. That means that the wording is bad.

If the problem had mentioned an unknown horizontal impulse at ##t=0##, all would have been well.

 

[pbuk]

That contradicts your second assertion that the object has no vertical velocity for ##t > 0##.

Where did I assert that? I asserted (or rather I repeated the statement in the question) that the object has no vertical velocity for t = 0.

 

[pbuk]

You cannot have it both ways. If being at rest at ##t=0## with a discontinuous change at ##t>0## is allowed for the horizontal velocity then surely the same can be true for the vertical velocity.

That would be the case if being at rest at ## t = 0 ## were the only condition, however there is another condition that vertical acceleration is constant, ruling out an initial discontinuity.

 

[PeroK]

Where did I assert that? I asserted (or rather I repeated the statement in the question) that the object has no vertical velocity for t = 0.

In your argument to rule out graph B?

Anyway, that fact that three senior HH are criticising this problem is enough in itself.

That would be the case if being at rest at ## t = 0 ## were the only condition, however there is another condition that vertical acceleration is constant, ruling out an initial discontinuity.

This is my objection. That's an arbirrary interpretation of kinematic data. A linear graph for horizontal displacement does not represent an object initially at rest. A linear graph represents an object with an initial horizontal component of velocity.

 

[jbriggs444]

That would be the case if being at rest at ## t = 0 ## were the only condition, however there is another condition that vertical acceleration is constant, ruling out an initial discontinuity.

Technically, that is not the case.

If the object is "at rest" at ##t=0## this means that we are contemplating times before the start of the problem. If we were only contemplating ##t \ge 0## then there would be no sense in which the object is at rest. Its horizontal velocity is constant at ##t>0##. Its [one-sided] horizontal velocity at ##t=0## is the same.

But you say that the object is initially at rest. So we must be contemplating negative times.

Velocity is the first derivative of displacement. It cannot change discontinuously from one constant value to another. The mean value theorem forbids it. So horizontal velocity is undefined at ##t=0##.

The same argument applied to vertical acceleration. It is a derivative. It cannot change discontinuously from one constant value to another. The mean value theorem forbids it. So vertical acceleration is undefined at ##t=0##.

So there is no initial discontinuity. Only a point at which both velocity and acceleration are undefined. Which makes the stipulation that the object begins at rest irrelevant.

 

[pbuk]

In your argument to rule out graph B?

Are you referring to this:

But you can't have a .... non-zero initial vertical component, and so B is not valid.

Here I talk about the initial velocity i.e. at ## t = 0 ##, not when ## t > 0 ##.

Anyway, that fact that three senior HH are criticising this problem is enough in itself.

Not all HH in this thread share your view that the problem is "a mess".

This is my objection. That's an arbirrary interpretation of kinematic data. A linear graph for horizontal displacement does not represent an object initially at rest. A linear graph represents an object with an initial horizontal component of velocity.

But there is no linear graph!