Ball Velocity at Different Times

[emma402]

A ball is thrown from the edge of a cliff with an initial velocity of 40·m/s upward. Answer the following questions using + (upward) and - (downward) to indicate the direction of the velocity. Ignore air resistance and assume the ball does not hit the ground during the question.
(a) Find the velocity of the ball 2 seconds after it is thrown. 1 m/s.

(b) Find the velocity of the ball 11 seconds after it is thrown. 2 m/s.

(c) How high does it rise above the edge of the cliff (from where it was thrown)? 3 m.

*need help! show work if possible!

 

[6Stang7]

You need to show your attempt at solving the problem first.

 

[emma402]

if the average velocity is change in position/elapsed time...then do I take 40/2=20 m/s? and the second one would be 40/11=3.66 m/s? for some reason I don't think that's right...and I don't even know where to begin on (C).

 

[6Stang7]

That would only work if the initial velocity was 40 m/s, and after 2 seconds the velocity was 0 m/s.

So, you throw a ball upward with some initial velocity and the ball, of course, goes upward because of this initial velocity. Other than the initial velocity, is there anything else acting on the ball?

 

[emma402]

the direction its going?

 

[6Stang7]

That's not really acting on the ball; that is the result of what is going on.

Say you drop a ball out of your hand, from rest. Why does the ball fall to the ground?

 

[emma402]

b.c of the force of gravity...so do i have to include the 10m/s pull from gravity? if sooo do i multiply 40m/s by 2s and divide it by 10m/s?

 

[6Stang7]

well, gravity is an acceleration (ie m/s^2).

I'm going to teach you a very valuable trick called dimensional analysis.

Ok, so your answer is what? Velocity. What are the units of velocity? m/s.

so, you want to do (40m/s)*(2s)/(10 m/s^2), but you don't know if that is right. You can check with dimensional analysis.

what you do is solve using just the units, so:

(m/s)*(s)/(m/(s^2))=(m/s)*(s)/(m/(s*s))=[(m*s*s*s)/(m*s)]=s*s=s^2

so that doesn't work (because you need m/s, not s^2).

So what do you need to do to get m/s?

 

[emma402]

do i use those 3 vaules?(40m/s,2s,10 m/s^2)

 

[6Stang7]

Yes, but the important question in how. Do you have a book that you can look in for equations?

 

[emma402]

ok so i looked up the equation (sorry I am teaching myself in this class so I know its agrivating helping me) and i used v=v0 + at...so for a i got 60m/s and 150m/s...i'm not sure how to start on C though. I have to figure out the direction of a and b...i think 60m/s is positive (upward) and 150m/s is neg. (downward)...but now I am having trouble with figuring out how hign the ball raises about the edge of the cliff from where it was thrown. It has to be in meters...so would I subtract the two numbers and multiply by the number of seconds?

 

[6Stang7]

ok so i looked up the equation (sorry I am teaching myself in this class so I know its agrivating helping me)

It's not aggravating in the slightest; as long as you're willing to work at it, we're more than eager to help :)

and i used v=v0 + at...so for a i got 60m/s and 150m/s...

That is the correct equation, but your answer are wrong; you made a simple mistake. Imagine in your head the process of you throwing a ball straight up in the air with some initial velocity. The ball is going to go up, stop, then go back down right? So, what happens to the velocity of the ball as it is going up? If up is positive and down is negative, and a in the equation v=v0 + at is gravity, what is the value of a?

i'm not sure how to start on C though. I have to figure out the direction of a and b...i think 60m/s is positive (upward) and 150m/s is neg. (downward)...but now I am having trouble with figuring out how hign the ball raises about the edge of the cliff from where it was thrown. It has to be in meters...so would I subtract the two numbers and multiply by the number of seconds?

The nice thing about the equation you have is that the value you get will tell you what way the ball is going. If the value is positive, which way is the ball going? What if the value is negative? Question C is asking how high the ball went, so you have velocity and acceleration, and time, and you need to solve for position. Can you find an equation in your book that has all those variables in it?


You're doing great and are really close to getting the correct answers for all three problems; keep it up :)

 

[emma402]

i don';t understand how my vaules are wrong? the math is done right isn't it? when you throw the ball up the velocity is going down or slowing down until it reaches 0 at the turning point.
the value of A=10m/s^2...(this is what we are using other than 9.8m/s)
if the value is pos it is going up and if neg it is going down. The equation I found with those values is the Velocity equation...if that's not right you might have to tell me what equation to use.

 

[6Stang7]

You mistake is with the value of a. Since up is positive and down is negative, and gravity acts down, that means a is...

 

[emma402]

ok so i think I'm solving for displacement then?...which would be the ave.velocity multiplied by time...and the ave velocity is 1/2(Vo + v)...so 1/2(40m/s + 60m/s)= 50m/s...if that is the ave. velocity then 50m/s(2s?) I'm confused on which time it wants...

 

[emma402]

ooo that means a= -10m/s^2?

after 2 sec. it would be a pos. 10m/s^2...after 11 sec that value would change to -10m/s^2 correct?

 

[emma402]

for (b) I now have a -70m/s for that velocity...i just wanted to see if that was right?

 

[6Stang7]

ooo that means a= -10m/s^2?

after 2 sec. it would be a pos. 10m/s^2...after 11 sec that value would change to -10m/s^2 correct?

SO CLOSE; you're almost there! a IS -10m/s^2, but it doesn't change its sign; it is always -10m/s^2. So why does the velocity of the ball go from negative to positive? You can see in the equation:

v=v0 + (-10m/s^2)t

that as t increases in value (as time goes forward) (-10m/s^2)t increases in value (it becomes a larger and larger negative number... when t=0, (-10m/s^2)t=0, when t=1, (-10m/s^2)t=-10m/s, when t=2, (-10m/s^2)t=-20m/s, and so on) until at some point, v=0. As time continues to go forward, (-10m/s^2)t keeps increasing in value (becoming a bigger and bigger negative number) and v becomes a negative number.

Knowing this, can you tell me at that point in time (value of t) does the ball reach 0 velocity?

 

[6Stang7]

for (b) i now have a -70m/s for that velocity...i just wanted to see if that was right?

bingo!

 

[emma402]

when t=0?

is (a) v= 60m/s?..b.c 2sec is still going up?

 

[6Stang7]

when t=0?

is (a) v= 60m/s?..b.c 2sec is still going up?

No on both. Your equation is v=v0 + at and when we plug in the given values we have

v=40m/s + (-10m/s^2)t and we want to find t when v=0. That is

0=40 + (-10m/s^2)t and solve for t. What do you get?


Can you write out your work for how you got 60 m/s^2?

 

[emma402]

you get 30m/s^2 right?

i took v=40m/s + (10m/s^2)(2s)=60m/s

"v=40m/s + (-10m/s^2)t and we want to find t when v=0. That is

0=40 + (-10m/s^2)t and solve for t. What do you get?"...i have no clue on the algebra on this!

 

[6Stang7]

Remember, a=-10m/s^2 and doesn't change.

so inside of v=40m/s + (10m/s^2)(2s) you should have v=40m/s + (-10m/s^2)(2s). What is that?

OK, so you have 0=40 + (-10m/s^2)t which is 40-10t=0

that is the same as 40=10t (I added 10t to both sides) which can be written as 40/10=t (I divided by 10 on both sides)which is?

It'd be a good idea to brush up on algebra; it is used a lot in physics

 

[emma402]

that is v=20m/s

that is 4, but what is that the answer to? the t in the displacement equation?

 

[6Stang7]

For t=2 seconds you are correct. The solving for t when the velocity is zero has nothing to do with the problem; I was hoping it'd help you understand what is going on.

For the final part, look in your book and see if you can find a equation that has distance, time, acceleration, and velocity as variables in it and post it.

 

[emma402]

i'm still wondering about the velocity for (a)...i feel like we were getting close and stopped working on that. can we work on that and then work on (c)?

for (a) is it v=20m/s??..that is what it comes out to be with the v=40m/s + (-10m/s^2)(2s)

and for (c) do I use the displacement equation or motion?

 

[emma402]

or did we answer (a) and I am not aware of it?