# Throwing a ball straight up

## Homework Statement

A ball is thrown straight up and it falls back to the ground.

## Homework Equations

Which one of the
following graphical representations describes the velocity of the ball as a
function of time.

## The Attempt at a Solution

I think the right answer is D, However i think it can also be C.
Tell me what you think!

#### Attachments

• ball.JPEG
15.6 KB · Views: 687

Doc Al
Mentor
How does the acceleration due to gravity relate to these diagrams?

tiny-tim
Homework Helper
I think the right answer is D, However i think it can also be C.
Tell me what you think!

well, it can't be both, can it? come on, chawki, you know the rules by now you tell us why you think it's D (or why it might be C), and then we'll comment Well, D because when we throw the ball, it will reach a time when the velocity will come to zero, and then it will fall, so basically there are two steps..
Step1: a1 < 0 and v1 > 0, a1*v1 < 0, so there is decceleration.
Step2: a2 < 0 and v2 < 0, a2*v2 > 0, so there is acceleration.

same thing for graph C

tiny-tim
Homework Helper
Well, D because when we throw the ball, it will reach a time when the velocity will come to zero, and then it will fall, so basically there are two steps..
Step1: a1 < 0 and v1 > 0, a1*v1 < 0, so there is decceleration.
Step2: a2 < 0 and v2 < 0, a2*v2 > 0, so there is acceleration.

same thing for graph C

(try using the X2 icon just above the Reply box )

i'm confused what is a1*v1 supposed to be? a1 is the acceleration during first step
v1 is the velocity during first step

tiny-tim
Homework Helper
but why are you multiplying them? Doc Al
Mentor
These diagrams show velocity as a function of time. Using them, how would you find the acceleration? Which diagram matches what you (should) know about the acceleration of an object in free fall?

but why are you multiplying them? It's a method i learnt at college

tiny-tim
Homework Helper
It's a method i learnt at college

explain These diagrams show velocity as a function of time. Using them, how would you find the acceleration? Which diagram matches what you (should) know about the acceleration of an object in free fall?

The acceleration is simply a1= difference of velocity during that step/difference of time during that step

explain a*v < 0 means there is decceleration
a*v > 0 means there is acceleration

tiny-tim
Homework Helper
a*v < 0 means there is decceleration
a*v > 0 means there is acceleration

hmm … i find that rather confusing …

by "acceleration" we usually mean acceleration in a particular direction (eg up), so that the acceleration of a projectile is always negative

so the test is:
a < 0 means there is negative acceleration
a > 0 means there is positive acceleration​

your method seems to be defining "acceleration" as meaning that the magnitude of the velocity increases (so a projectile always decelerates on the way up and always accelerates on the way down)

i don't like your method at all, it's so complicated

surely a < 0 is an easier test than a*v < 0 ? oh, and i still don't understand why, even on your test, you think C might be right Yes C might be right too because when we apply that method which you don't like :D we find same thing as D
one other thing..i learnt in french..maybe they gave us different definitions..all i know and remember is that it works.
so what do you think C or D :D

Doc Al
Mentor
The acceleration is simply a1= difference of velocity during that step/difference of time during that step
Exactly right. So how do the accelerations exhibited by diagrams C and D differ? Hint: Pay attention to signs.

if we take the method of looking only to the acceleration, i think the answer would be C only.

Doc Al
Mentor
if we take the method of looking only to the acceleration, i think the answer would be C only.
Why do you say that?

What property of a velocity versus time graph tells you the acceleration?

Realize that acceleration is a vector and thus has direction (and thus sign).

yes yes, i'm aware.
i just used to use acceleration*velocity to see if it's acceleration or deceleration.
maybe i was wrong???

tiny-tim
Homework Helper
hi chawki! now i've slept on it :zzz:, i can see why your method is giving you C as an alternative answer

when the projectile comes down, it is accelerating (ie the magnitude of the velocity is increasing), so your method requires a*v to be positive

in the graph C, a and v are both positive, so from that point of view C is consistent with that

(and so is D, the correct answer, with both a and v negative)

unfortunately, C corresponds to half a parabola on the way up, then the other half of the parabola upside down, with the projectile accelerating off into outer space (so that the whole thing looks a bit like a y = x3 graph)

(can you se that now?)

i hope you now see that this acceleration/deceleration method, using the sign of both a and v, is potentially misleading, and it's best just to talk about positive and negative acceleration, and use only the sign of a Ok Tiny-tim,
it's a good point to clear up this confusing matter, i will use only 'a' to see if there is acceleration or deceleration, with a = difference in velocity/ difference in time.
and then the answer would be only C.

tiny-tim
Homework Helper
… i will use only 'a' to see if there is acceleration or deceleration, with a = difference in velocity/ difference in time.
and then the answer would be only C.

why C ? in which direction are you measuring v and a ?

because
in step 1
a1= (0-v)/(t1-t0) and we get a negative value.
in step2
a2= (v-0)/(t2-t1) = positive value.

so it makes sense.

Doc Al
Mentor
because
in step 1
a1= (0-v)/(t1-t0) and we get a negative value.
in step2
a2= (v-0)/(t2-t1) = positive value.

so it makes sense.
What's step 1 and step 2? Are you still looking at diagram C?

Is the acceleration of a falling body sometimes positive and sometimes negative? (Which way does gravity act?)

tiny-tim
Homework Helper
because
in step 1
a1= (0-v)/(t1-t0) and we get a negative value.
in step2
a2= (v-0)/(t2-t1) = positive value.

so it makes sense.

chawki, what you are saying (correctly) is that in step 1 of graph C, the acceleration upward is negative, and in step 2 of graph C, the acceleration upward is positive …

as Doc Al is hinting, do you really think that is correct for a projectile under gravity? I don't mean to confuse chawki more, but is that graph a good representation? If they mean the size of the vector with v, can't it be C then? The size only isn't sign dependant right? Or do they mean the z-component of the velocity vector?