Velocity of a Ball Thrown from a Building

[chawki]

i think it's -12 in the case of free fall ?

 

[Doc Al]

i think it's -12 in the case of free fall ?

"Free fall" just means that the only force acting is gravity. However you throw the ball--up, down, or sideways--once it leaves your hand it's in free fall. (Ignoring air resistance, of course.)

 

[chawki]

That's confusing :(
in another problem, there was an object at the roof of a building, then it falls..
they asked the time it takes the object to reach the ground..and if i use (as you told me)
h=-1/2*V*t^2 ..i won't get t...
my mind is messed now

 

[chawki]

and same thing if i want to find the velocity when that object hits the ground...

 

[Doc Al]

That's confusing :(
in another problem, there was an object at the roof of a building, then it falls..
they asked the time it takes the object to reach the ground..and if i use (as you told me)
h=-1/2*V*t^2 ..i won't get t...
my mind is messed now

To be consistent, you should use:
Δh = -1/2gt^2

Works fine. What would you use for Δh?

(If you let down be positive, then you can use h = 1/2gt^2. But let's get the standard convention straight before changing things.)

 

[Doc Al]

and same thing if i want to find the velocity when that object hits the ground...

What equation are you using?

 

[chawki]

To be consistent, you should use:
Δh = -1/2gt^2

Works fine. What would you use for Δh?

(If you let down be positive, then you can use h = 1/2gt^2. But let's get the standard convention straight before changing things.)

Δh=final distance-initial distance
but since you talked about the possibility of writing h = 1/2gt^2 , please please please tell me what i should change in the equations that i wrote in post #30
i prefer using down as positive direction

 

[Doc Al]

i prefer using down as positive direction

Why?

 

[cupid.callin]

Why?

i guess that way he doesn't have to worry about sign of g

i also used to do it when i started studying physics

 

[chawki]

Why?

i just prefer...i used to it when i was at school...and i have an important exam to take..and not enough time to revise everything..and using the down direction as negative will just add more stress on me and confuse me more

 

[chawki]

i guess that way he doesn't have to worry about sign of g

i also used to do it when i started studying physics

yes ..thank you...you are right...i don't want to take care about the sign of g or V...i just want the correct equations for ''down as positive''

 

[Doc Al]

i guess that way he doesn't have to worry about sign of g

If you use a standard sign convention, you never have to worry about the direction of gravity.

i also used to do it when i started studying physics

Sure, me too. But I strongly suggest that one learn the standard method that applies to all problems. Once you've mastered that, then you can take short cuts.

 

[Doc Al]

yes ..thank you...you are right...i don't want to take care about the sign of g or V...i just want the correct equations for ''down as positive''

Here's how to figure it out:

Start with the general equations for uniformly accelerated motion. You can find them here: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

To apply them to free fall with down = positive, use a = g.
To apply them to free fall with up = positive, use a = -g.

Done!

(But don't be surprised if you mess things up, since now you have to remember multiple equations. The 'down as positive' method is fine when nothing changes direction, such as when you throw something downward. Not so good when you throw things up, since they then come down.)

 

[chawki]

If you use a standard sign convention, you never have to worry about the direction of gravity.

Sure, me too. But I strongly suggest that one learn the standard method that applies to all problems. Once you've mastered that, then you can take short cuts.

i know what you mean...but I DON'T HAVE TIME :(

 

[chawki]

cupid.callin please HELLP ME if you have the equation for both cases (free falls, throwing downward) and (throwing upward)

 

[Doc Al]

Here you go:

Let me get this right,
in case of falls and throwing down with initial speed, the equations are:
h=1/2*g*t²+V₀*t
V=g*t+V₀
V²-V₀²=-2*g*Δh ?

in case of throwing the ball upward, the equations are:
h=-1/2*g*t²+V₀*t
V=-g*t+V₀
V²-V₀²=-2*g*Δh ?

Here are the equations if up = positive (a = -g):
y = y₀ + v₀t - 1/2gt^2
(or Δh = v₀t - 1/2gt^2)
v = v₀ - gt
V²-V₀² = -2*g*Δh

If you prefer down = positive, just change the sign of g (a = +g):
Here are the equations if up = positive:
y = y₀ + v₀t + 1/2gt^2
(or Δh = v₀t + 1/2gt^2)
v = v₀ + gt
V²-V₀² = 2*g*Δh

 

[cupid.callin]

DOC AL is helping you and you are asking for my help ...

He is far more intelligent that both of us combined (Thats why he's a mentor ... LOL)

Anyway ...

these 3 eqn's will solve all your problems (provided you use them correctly)

v = u + at
s = ut + 0.5at²
v² = u² + 2as

and always remember that v,u,a,s (and s is displacement not distance)are all vectors (not used as vectors there though)

do it simple and step by step

lets do a question ...

suppose you throw a ball up with speed 50 m/s ... what will be its speed and height after 2 sec

(use appropriate eqn and solve)

 

[chawki]

DOC AL is helping you and you are asking for my help ...

He is far more intelligent that both of us combined (Thats why he's a mentor ... LOL)

Anyway ...

these 3 eqn's will solve all your problems (provided you use them correctly)

v = u + at
s = ut + 0.5at²
v² = u² + 2as

and always remember that v,u,a,s (and s is displacement not distance)are all vectors (not used as vectors there though)

do it simple and step by step

lets do a question ...

suppose you throw a ball up with speed 50 m/s ... what will be its speed and height after 2 sec

(use appropriate eqn and solve)

y=1/2*g*t²+V₀*t = 119.62 m
V=g*t+V₀=69.62 m/s

 

[cupid.callin]

y=1/2*g*t²+V₀*t = 119.62 m
V=g*t+V₀=69.62 m/s

So you think if i launch something with m/s upwards .. it will increase its speed?

you you can go into space by just jumping ...?

Imagine the problem ... how it will proceed

and if v_o is positive then g will be ?

 

[chawki]

So you think if i launch something with m/s upwards .. it will increase its speed?

you you can go into space by just jumping ...?

Imagine the problem ... how it will proceed

and if v_o is positive then g will be ?

i think that if v_o is positive then g will be negative
y=-1/2*g*t²+V₀*t = 80.38 m
V=-gt+v₀ = 30.38 m/s

 

[cupid.callin]

yes now its correct