Determine initial velocity of a vertical throw

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Homework Help Overview

The problem involves a ball thrown vertically upward, returning to its initial position after 4 seconds, with gravity set at 10 m/s². Participants are exploring the initial velocity required for this motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to use kinematic equations, such as \(v = v_0 + at\) and \(s = v_0t + \frac{1}{2}at^2\), but express uncertainty about the implications of time and acceleration during the upward and downward phases of the motion.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the assumptions about time and acceleration. Some guidance has been offered regarding the relationship between upward and downward motion, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are grappling with the concept of whether the time taken to ascend is equal to the time taken to descend, which affects their calculations. There is also confusion regarding the velocity of the ball at the moment it returns to the starting height.

duchuy
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Homework Statement
t = 4s , g = 10 m/s^2
Relevant Equations
I don't really know which equation to use...
Hi,
I was given this problem saying that a ball is thrown vertically up in the air and returns to its initial position after 4 seconds. The acceleration due to gravity is given to be equal to 10m/s^2.
I tried to attempt this problem by using the equation :
v^2 - v0^2 = 2ah by considering terminal velocity = 0. Despite this, i still can't determine the total distance travelled.
Sorry in advance if I have misused any vocabulary sincei translated this from french.
Thank you!
 
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Do you have any ideas?
 
PeroK said:
Do you have any ideas?
Well, I'm also trying with v = v0 + at but I didnt really work out..
This is why I chose medecine physics is so so hard..
 
duchuy said:
Well, I'm also trying with v = v0 + at but I didnt really work out..
What about $$s = v_0t + \frac 1 2 at^2$$
 
PeroK said:
What about $$s = v_0t + \frac 1 2 at^2$$
But in this case wouldn't I have to values of gravitational acceleration?
When the object is going up, acceleration is negative and when it goes down, it's positive.
Therefore I couldn't use the 4 seconds value because ( i think ), the time the object takes to go up is different from the time it takes to go down right?
 
duchuy said:
Well, I'm also trying with v = v0 + at but I didnt really work out..
This is why I chose medecine physics is so so hard..
That equation is a good start. You know that the ball is in the air for 4 s, and you know the acceleration. If the velocity is +v0 when the ball is thrown up, what would it be at the moment it comes back down after 4 s?
 
kuruman said:
That equation is a good start. You know that the ball is in the air for 4 s, and you know the acceleration. If the velocity is +v0 when the ball is thrown up, what would it be at the moment it comes back down after 4 s?
I think the thing that what's stopping me is that I don't know whether the time an object takes to go up is different from the time it takes to go down..
Because I've been considering that it's different, so i didnt really manage to write the equation..
At 4s, wouldn't the velocity be 0?
If I consider that the time is equal, I'm getting v0 = 20 m/s ...
 
duchuy said:
But in this case wouldn't I have to values of gravitational acceleration?
When the object is going up, acceleration is negative and when it goes down, it's positive.
Therefore I couldn't use the 4 seconds value because ( i think ), the time the object takes to go up is different from the time it takes to go down right?
If you're not sure, you could split the motion into two parts (up and down). You can then show using the kinematic equations whether the time to go up is the same as the time to come down.
 
duchuy said:
At 4s, wouldn't the velocity be 0?
If I consider that the time is equal, I'm getting v0 = 20 m/s ...
If you throw a ball up at ##20 m/s##:

1) What is the displacement after ##4s##?

2) What is the velocity after ##4s##?
 
  • #10
duchuy said:
I think the thing that what's stopping me is that I don't know whether the time an object takes to go up is different from the time it takes to go down..
Because I've been considering that it's different, so i didnt really manage to write the equation..
At 4s, wouldn't the velocity be 0?
If I consider that the time is equal, I'm getting v0 = 20 m/s ...
At 4 s the ball returns to the height from which it was launched. Just before it is stopped, it is still moving, That's the velocity I was asking about. Anyway, I don't want to detract from the course that @PeroK has set so I will cease and desist.
 
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  • #11
PeroK said:
If you throw a ball up at ##20 m/s##:

1) What is the displacement after ##4s##?

2) What is the velocity after ##4s##?
Oh ok ok I got that thank you two so much!
For the displacement after 40swouldn't it be 40m?
For the velocity I'm not quite sure.. I really thought that it would be 0 since it's the moment the ball landed..
 
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  • #12
duchuy said:
Oh ok ok I got that thank you two so much!
For the displacement after 40swouldn't it be 40m?
For the velocity I'm not quite sure.. I really thought that it would be 0 since it's the moment the ball landed..
Sorry, that's not right at all.

1) That's not what the equations tell you.

2) A ball is moving downwards when it hits the ground after being thrown up.

There are lots of videos on line about throwing an object up. I think you need to develop some sort of physical understanding of what is happening.

But, also, you have to develop the ability to use equations.
 

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