Opposing accelerations on stone drop/throw up problem

[Tallgirl]

Homework Statement

A stone is dropped from the top of a 50m high tower at the same moment as a ball is thrown straight upward from the bottom of the tower, with an initial velocity of 20m/s. when and where do the ball and stone pass each other?

Homework Equations

I actually have the solution in front of me. My frustration comes when it identies the given variables. The acceleration of the thrown ball is negative as the velocity is positive. This means that down is negative. In the solution, the dropped stone is assigned a positive acceleration. To me the stone is speeding up in the negative direction, isn't that negative acceleration? How can each object have separate direction's considering they are both used in the solution?

The Attempt at a Solution

 

[SammyS]

Homework Statement

A stone is dropped from the top of a 50m high tower at the same moment as a ball is thrown straight upward from the bottom of the tower, with an initial velocity of 20m/s. when and where do the ball and stone pass each other?

Homework Equations

I actually have the solution in front of me. My frustration comes when it identies the given variables. The acceleration of the thrown ball is negative as the velocity is positive. This means that down is negative. In the solution, the dropped stone is assigned a positive acceleration. To me the stone is speeding up in the negative direction, isn't that negative acceleration? How can each object have separate direction's considering they are both used in the solution?

The Attempt at a Solution

Hello Tallgirl. Welcome to PF !

Can you give the two equations, exactly as they're written in that solution?

 

[Sunil Simha]

I actually have the solution in front of me. My frustration comes when it identies the given variables. The acceleration of the thrown ball is negative as the velocity is positive. This means that down is negative. In the solution, the dropped stone is assigned a positive acceleration. To me the stone is speeding up in the negative direction, isn't that negative acceleration? How can each object have separate direction's considering they are both used in the solution?

Try following your method (stick to the sign convention) and see whether you get a different answer.

(Note: If you take the downward direction as negative then all vectors pointing downwards, including displacement will be negative)

 

[Sunil Simha]

Did you get the equations?

 

[Tallgirl]

Here's how it is given:

Stone's variables:
vi = 0m/s
a = 9.8m/s*2
t = ?
d = .5at*2 = (4.9m/s*2)(t*2)

Ball's variables:
vi = +20m/s
a = -9.8m/s*2
t = ?
d = vit + .5at*2 = (+20m/s)t + (-4.9m/s*2)(t*2)

d1 + d2 = 50m

(4.9m/s*2)(t*2) + (+20m/s)t + (-4.9m/s*2)(t*2) = 50m

(+20m/s)t = 50m

t =2.5s

And then plug this into the d equations to find the distances.

 

[Sunil Simha]

Here's how it is given:

Stone's variables:
vi = 0m/s
a = 9.8m/s*2
t = ?
d = .5at*2 = (4.9m/s*2)(t*2)

d1 + d2 = 50m

This is correct.

But according to your sign convention, a = -9.8 m/s², right? This would then mean, the d that you have found (0.5gt²) will also be negative. But then, d₁+d₂ wouldn't be the same right?

This seemingly paradoxical statement is a result of not taking proper signs into consideration.

Now if total distance is D (from the bottom) and d is the displacement of the stone from the high point, d would be negative. Then the displacement of the stone from the bottom would be D+d( vector sum;that means, you are taking them along with the signs). Its magnitude would be D-d and hence when the stone and ball meet, (with d₂ being the ball's displacement from the bottom) D= d₂ - d (vector difference). Thus since d is negative, magnitude of d₂ - d is d₂ + d which is D.

 

[Tallgirl]

Thank you so much.