Velocity Problem: Ball Thrown Upwards, +30 m/s, After 4s

[fattydq]

A ball is thrown upwards with a velocity of +30 m/s. At the end of 4 s, its velocity will be closest to
A) -50 m/s
B) -20 m/s
C) -10 m/s
D) +10 m/s

I really don't know what equations would be involved to solve this problem. I'm thinking the downward force of gravity can be taken into account or something like that?

 

[Fightfish]

I really don't know what equations would be involved to solve this problem. I'm thinking the downward force of gravity can be taken into account or something like that?

That would be the kinematics equations, one of which is v = u + at.

 

[fattydq]

That would be the kinematics equations, one of which is v = u + at.

But what does u represent here? and would the acceleration just be 30??

 

[kuruman]

What is your understanding of acceleration? Specifically, how does acceleration relate to velocity?

 

[fattydq]

Is acceleration not just the rate that velocity is changing?

 

[fattydq]

I don't understand what that has to do with my question though...although I'm sure you plan on explaining, what does this have to do with figuring out what equation to use to solve this problem?

 

[kuruman]

Yes, it is. So if the acceleration of gravity is -9.8 m/s², this means that for every second that goes by, you add -9.8 m/s to the velocity that is already there. Look at your question again. At time t = 0 the velocity that is already there is +30 m/s. What must you add to it after 4 s? What do you get for velocity when you do that?

 

[kuruman]

I don't understand what that has to do with my question though...although I'm sure you plan on explaining, what does this have to do with figuring out what equation to use to solve this problem?

What I described in posting#7 is the equation

v = v₀ - g*t

 

[fattydq]

Yes, it is. So if the acceleration of gravity is -9.8 m/s², this means that for every second that goes by, you add -9.8 m/s to the velocity that is already there. Look at your question again. At time t = 0 the velocity that is already there is +30 m/s. What must you add to it after 4 s? What do you get for velocity when you do that?

So after four seconds you would add (4x-9.8) to thirty, correct?

 

[fattydq]

Thus my answer would be D...since I'm still going to have a positive number. Right?

 

[kuruman]

So after four seconds you would add (4x-9.8) to thirty, correct?

4*(-9.8) = -39.2 m.s Add that to 30 and you get ...

 

[fattydq]

Ahh, yes, duh. Haha, it seems very simple now...but I still don't understand what this v=u+at means?

 

[kuruman]

Ahh, yes, duh. Haha, it seems very simple now...but I still don't understand what this v=u+at means?

Look at posting#8. replace u with v₀ and a with -g. What do you get? The expression v = u + at is a general form of the same equation.

 

[Fightfish]

Ahh, yes, duh. Haha, it seems very simple now...but I still don't understand what this v=u+at means?

It means that for a body undergoing constant acceleration, the final velocity v of the body after a time t is equivalent to the initial velocity u of the body plus the change in the velocity of the body during this time (ie acceleration X time), essentially what you've been doing just now. *Take note of the directions when using this equation, all the terms are vector quantities.*

 

[tiny-tim]

Ahh, yes, duh. Haha, it seems very simple now...but I still don't understand what this v=u+at means?

Hi fattydq!

Think of it as v - u = at, so a = (v - u)/t …

then it simply says that acceleration is change-in-speed divided by time… which it is, isn't it?

( of course, this only works for constant acceleration )