Opposing accelerations on stone drop/throw up problem

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In summary: I really appreciate your help.In summary, the problem involves a stone being dropped from a 50m high tower and a ball being thrown upwards from the bottom of the tower with an initial velocity of 20m/s. The question is when and where the ball and stone will pass each other. The solution involves using equations for displacement, velocity, and acceleration for both objects, taking into account their different directions. The solution shows that the objects will pass each other after 2.5 seconds, with the stone having a negative displacement and the ball having a positive displacement.
  • #1
Tallgirl
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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

 
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  • #2
Tallgirl said:

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?
 
  • #3
Tallgirl said:
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)
 
  • #4
Did you get the equations?
 
  • #5
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.
 
  • #6
Tallgirl said:
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.
 
  • #7
Thank you so much.
 

1. What is the concept of opposing accelerations in the stone drop/throw up problem?

In the stone drop/throw up problem, opposing accelerations refer to the forces acting on an object in opposite directions that affect its motion. In this scenario, the force of gravity pulls the object downwards while the force of air resistance pushes the object upwards.

2. How do opposing accelerations affect the velocity of the object?

The opposing accelerations cause the object to have a constantly changing velocity. As the object is dropped or thrown upwards, the force of gravity causes it to accelerate downwards while the force of air resistance causes it to decelerate. This results in a decrease in velocity until the object reaches its maximum height, where the forces are balanced and the velocity is momentarily zero. As the object falls back down, the forces cause it to accelerate downwards again, increasing its velocity until it reaches the ground.

3. What happens to the acceleration of the object during its motion?

The acceleration of the object changes throughout its motion due to the opposing forces acting on it. Initially, the acceleration is positive as the object is thrown or dropped, but as it moves upwards, the acceleration decreases until it reaches zero at its maximum height. As the object falls back down, the acceleration becomes negative as it is pulled downwards by gravity.

4. How does the mass of the object affect the opposing accelerations?

The mass of the object does not directly affect the opposing accelerations. However, a larger mass will experience a greater force of gravity, resulting in a greater acceleration downwards. This can also affect the force of air resistance, as a larger object will experience more air resistance, causing a greater deceleration as it moves upwards.

5. How can the stone drop/throw up problem be solved using equations?

The stone drop/throw up problem can be solved using the equations of motion, specifically the equations for velocity and acceleration. These equations take into account the initial velocity of the object, the acceleration due to gravity, and the time elapsed. By plugging in the known values and solving for the unknown, the motion of the object can be determined at any given time during its trajectory.

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