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chawki
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Homework Statement
A ball is thrown up from the top of a 48 m tall building with an initial velocity of 12 m/s.
Homework Equations
What is the velocity of the ball at the ground?
The Attempt at a Solution
V=-9.81*t+12 ?
That's the velocity as a function of time. What about the position?chawki said:V=-9.81*t+12 ?
cupid.callin said:just use v2 = u2 + 2as
chawki said:V2=(-2gh)+V02
V=28.24 m/s ?
That's fine. (Assuming you use the correct value for h, which really should be Δh.)chawki said:V2=(-2gh)+V02
But you messed up somewhere. Show what you plugged in where.V=28.24 m/s ?
That equation doesn't depend on the direction in which you throw. The negative sign is correct because the acceleration is -g (taking down as negative). But what's 'h'? That's where sign is also important.chawki said:it shouldn't be V2-V02= - 2*g*h ? since we are throwing upward.
and we don't know how much the ball got higher before falling
tiny-tim said:no
isn't -2gh positive ?
Doc Al said:That's fine. (Assuming you use the correct value for h, which really should be Δh.)
But you messed up somewhere. Show what you plugged in where.
chawki said:V= [tex]\sqrt{}-2*9.81*-48+12^2[/tex]
V= 32.95 m/s
Good.chawki said:i understand Δh=-48m ?
That's a different problem, but you can use the same equation to solve it. In that case you'd be solving for Δh.( but still i wonder how much the ball got higher when we throw it upward)
Doc Al said:Good.
That's a different problem, but you can use the same equation to solve it. In that case you'd be solving for Δh.
That's right. When the ball comes back down to its initial level, it's going at the same speed it started out at. We don't need to know how high it went.chawki said:ok, so i understand it doesn't change anything in our solution how much the ball got higher? we use only the height from the initial throw to the ground ?
Doc Al said:That's right. When the ball comes back down to its initial level, it's going at the same speed it started out at. We don't need to know how high it went.
Right. Whether you throw the ball up or down, you'll get the same speed when it reaches the ground. Obviously throwing it up means its trip to the ground takes longer.chawki said:so it's like we throwed the ball downward with an initial speed?
No. The equation doesn't change. It's the same regardless of the direction you throw.and in that case..wouldn't be V2-V02 = 2*g*h ?
That's because you made the acceleration positive. But gravity still acts down.with h=-48m
and then we have a problem..we can't find the root of a negative number...
Doc Al said:Right. Whether you throw the ball up or down, you'll get the same speed when it reaches the ground. Obviously throwing it up means its trip to the ground takes longer.
No. The equation doesn't change. It's the same regardless of the direction you throw.
That's because you made the acceleration positive. But gravity still acts down.
That equation should be:chawki said:I have learned that the equation of throwing a ball downward is typicall to free fall eauation..which is: V2-V02 = 2*g*h
Doc Al said:That equation should be:
V2-V02 = -2*g*Δh
The minus sign is important. (a = -g = -9.8 m/s^2)
Yes.chawki said:are you saying that in case of free fall, we write: V2-V02 = -2*g*Δh
:rofl: Yeah, not the best title choice.cristo said:<offtopic post>
I had the change the title: I couldn't keep reading the title "throwing up"!
Doc Al said::rofl: Yeah, not the best title choice.
cristo said:I had the change the title: I couldn't keep reading the title "throwing up"!
cupid.callin said:v2-u2 = 2gΔh is the eqn in its pure form ...
to use it ... just choose anyone direction (up or down) as positive(just like Y axis) and other as negative(obvious)
and then write the values with signs corresponding to your direction ...
like ... if you choose up as positive then eqn would be ...
v2 - 122 = 2(-9.8)(-48)
v2 - 122 = 2(9.8)(48)
thus v = 32.95 m/s
and in case you throw the ball down ... (taking up as positive)
v2 - (-12)2 = 2(-9.8)(-48)
v2 - 122 = 2(9.8)(48)
so you get same answer
... and this is in very obvious ...
when you throw the ball up ... it goes to max height and then return to the thrower's level having same speed 12 but now downwards ... so obviously of you initially throw the ball downwards ... it will still gain same speed when it reached the bottom
No. The equations should be the same for both cases. The acceleration is always downward, a = -g.chawki said: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 ?
"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 said:i think it's -12 in the case of free fall ?
To be consistent, you should use:chawki said: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