Determine initial velocity of a vertical throw

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SUMMARY

The discussion focuses on determining the initial velocity of a ball thrown vertically upward, which returns to its starting position after 4 seconds, with gravitational acceleration set at 10 m/s². Participants explore various kinematic equations, including s = v0t + 1/2 at² and v = v0 + at, to calculate the initial velocity. The consensus is that the total time of flight is equally divided between ascent and descent, leading to an initial velocity of 20 m/s. Misunderstandings regarding the velocity at the peak and displacement are clarified, emphasizing the importance of correctly applying kinematic principles.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Basic knowledge of gravitational acceleration (10 m/s²)
  • Familiarity with concepts of velocity and displacement
  • Ability to differentiate between upward and downward motion
NEXT STEPS
  • Study the derivation and application of kinematic equations
  • Learn about the concept of terminal velocity and its implications
  • Explore video resources on projectile motion for visual understanding
  • Practice problems involving vertical throws and their kinematic analysis
USEFUL FOR

Students in physics, educators teaching kinematics, and anyone interested in understanding the principles of vertical motion and projectile dynamics.

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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