Velocity of a Ball Thrown from a Building

[chawki]

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 ?

 

[tiny-tim]

hi chawki!

you need one of the other constant acceleration equations …

yours uses v_i v_f a and t, but you don't know t

you need one that uses v_i v_f a and s ​

 

[cupid.callin]

just use v² = u² + 2as

 

[Doc Al]

V=-9.81*t+12 ?

That's the velocity as a function of time. What about the position?

 

[chawki]

just use v² = u² + 2as

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

 

[cupid.callin]

i gave you the eqn in original ... you should the direction yourself

 

[chawki]

V²=(-2gh)+V₀²
V=28.24 m/s ?

 

[tiny-tim]

V²=(-2gh)+V₀²
V=28.24 m/s ?

no

isn't -2gh positive ?

 

[Doc Al]

V²=(-2gh)+V₀²

That's fine. (Assuming you use the correct value for h, which really should be Δh.)

V=28.24 m/s ?

But you messed up somewhere. Show what you plugged in where.

 

[Doc Al]

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

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.

 

[cupid.callin]

v² and u² will be positive no matter what


what do you think will be signs of s and a ?

 

[chawki]

no

isn't -2gh positive ?

 

[chawki]

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.

i understand Δh=-48m ? ( but still i wonder how much the ball got higher when we throw it upward)

well..if Δh=-48
V= \sqrt{}-2*9.81*-48+12^2
V= 32.95 m/s

 

[cupid.callin]

V= \sqrt{}-2*9.81*-48+12^2
V= 32.95 m/s

Yes its right

And next time wry writing "-2*9.81*-48+12^2" inside the bracket of sqrt in latex ... LOL ...

 

[Doc Al]

i understand Δh=-48m ?

Good.

( but still i wonder how much the ball got higher when we throw it upward)

That's a different problem, but you can use the same equation to solve it. In that case you'd be solving for Δh.

 

[chawki]

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.

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]

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 ?

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]

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.

That's very interesting.. Thank you so much
so it's like we throwed the ball downward with an initial speed?
and in that case..wouldn't be V²-V₀² = 2*g*h ?
with h=-48m
and then we have a problem..we can't find the root of a negative number...

 

[Doc Al]

so it's like we throwed the ball downward with an initial speed?

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.

and in that case..wouldn't be V²-V₀² = 2*g*h ?

No. The equation doesn't change. It's the same regardless of the direction you throw.

with h=-48m
and then we have a problem..we can't find the root of a negative number...

That's because you made the acceleration positive. But gravity still acts down.

 

[chawki]

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.

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]

I have learned that the equation of throwing a ball downward is typicall to free fall eauation..which is: V²-V₀² = 2*g*h

That equation should be:
V²-V₀² = -2*g*Δh

The minus sign is important. (a = -g = -9.8 m/s^2)

 

[cristo]

<offtopic post>

I had the change the title: I couldn't keep reading the title "throwing up"!

 

[chawki]

That equation should be:
V²-V₀² = -2*g*Δh

The minus sign is important. (a = -g = -9.8 m/s^2)

are you saying that in case of free fall, we write: V²-V₀² = -2*g*Δh

 

[Doc Al]

are you saying that in case of free fall, we write: V²-V₀² = -2*g*Δh

Yes.

 

[Doc Al]

<offtopic post>

I had the change the title: I couldn't keep reading the title "throwing up"!

Yeah, not the best title choice.

 

[chawki]

Yeah, not the best title choice.

sorry

 

[tiny-tim]

I had the change the title: I couldn't keep reading the title "throwing up"!

I'd have changed it to "throwing up (sic)" …

and then washed my hands of it! ​

 

[cupid.callin]

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

 

[chawki]

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

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 ?

 

[Doc Al]

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 ?

No. The equations should be the same for both cases. The acceleration is always downward, a = -g.

The only difference between throwing the ball upwards versus throwing it downwards is the sign of V₀: In one case, V₀ = +12 m/s (for example); in the other, V₀ = -12 m/s.