Calculating the velocity and acceleration given vector velocity at 1 second

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Homework Help Overview

The problem involves a physics student analyzing the motion of a ball thrown into the air on Mars, focusing on its velocity and acceleration at specific time intervals. The ball follows a parabolic trajectory, reaching maximum height at 2 seconds, with given velocity components at 1 second.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the velocity of the ball at maximum height and question the implications of the gravitational acceleration on the motion. There are attempts to clarify the relationship between velocity and acceleration, particularly at the peak of the trajectory.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem. Some have provided guidance on understanding the components of velocity and the effects of gravity, while others are questioning assumptions about the motion of the ball.

Contextual Notes

There is some confusion regarding the reference to the moon in the problem statement, with participants suggesting it should refer to Mars. Additionally, the problem constraints include the assumption of no air resistance.

mdavies23

Homework Statement


A physics student studying gravity on the surface of Mars, throws a ball into the air at time t=0s. The ball follows a parabolic trajectory as shown and reaches a maximum height height at t=2s. At t=1s the velocity of the ball is v(t=1s)= 2.00 m/s i + 3.72 m/s j where i and j are unit vectors in the x and y directons. You may ignore "air" resistance.

a.) Write the balls velocity in vector form at t = 2s and t = 3s
b.) From the information provided, calculate the magnitude of acceleration of gravity near the surface of the moon.
c.) Determine the initial speed and launch angle.
d.) How far away from the launch point does the ball land.

Homework Equations


sqrt(a^2 + b^2)
tan^-1(y/x)

The Attempt at a Solution


Honestly, I do not know how to start. I have drawn the velocity vectors and the graph of the ball.

sqrt(2^2 + 3.72^2) = 4.22 m/s
tan^-1(3.72/2) = 61.73 degrees
 
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Start from part (a). If the ball reaches max. height in 2 s, what is its velocity in vector form, i.e. what are the vertical and horizontal components when t = 2 s?

On edit: Part (b) asks for the acceleration of gravity near the surface of the moon. I am sure it means "Mars".
 
If the gravitational acceleration is gM, how does the vertical velcity vary with time?

I presume the reference to moon in part b is a mistake.
 
kuruman said:
Start from part (a). If the ball reaches max. height in 2 s, what is its velocity in vector form, i.e. what are the vertical and horizontal components when t = 2 s?

On edit: Part (b) asks for the acceleration of gravity near the surface of the moon. I am sure it means "Mars".
So would that just be 2*2 m/s i + 3.72*2 m/s j?
 
What does maximum height mean?
 
kuruman said:
What does maximum height mean?
velocity equals 0
 
That's when the ball is shot straight up. This is not the case here. Think again, this an important point for you to understand.
 
kuruman said:
That's when the ball is shot straight up. This is not the case here. Think again, this an important point for you to understand.
It is when the ball starts coming down. Why wouldn’t y velocity be 0
 
Because the ball moves in a parabolic trajectory. When it's at maximum height, it is still moving parallel to the ground. If the velocity were zero at max height, the ball would drop straight down once it reaches the top of its trajectory. That's not what happens. Throw a ball across the room and you will see what I mean.
 
  • #10
kuruman said:
Because the ball moves in a parabolic trajectory. When it's at maximum height, it is still moving parallel to the ground. If the velocity were zero at max height, the ball would drop straight down once it reaches the top of its trajectory. That's not what happens. Throw a ball across the room and you will see what I mean.
True, so would acceleration be 0
 
  • #11
You are perhaps confusing acceleration and velocity. If the velocity is zero and the acceleration is zero, then the velocity will not change which means that the ball will hang at maximum height forever. This is not what happens. I'll give you a hint: at maximum height the ball instantaneously is not rising and is not dropping. Something is zero, but what?
 
  • #12
I guess I do'nt understand. I would think the Vy would be zero? The velocity vector is parallel to the x-axis at that instance.
 
  • #13
mdavies23 said:
I would think the Vy would be zero?
That is correct, vy is instantaneously zero. It starts at v0y then is reduced to zero because every second that goes by, g m/s are added in the "down" direction to the vy that is already there. That's what the equation vy = v0y - gt says in the language of mathematics. What about vx? How does that change as time goes by?
 
  • #14
It would be a decreasing straight line
 
  • #15
If the velocity in the horizontal direction is decreasing, this means that there is a horizontal acceleration. Perform this simple experiment that I am sure you done many times. Hold a ball between your thumb and forefinger and just let go. I am sure you know it that will simply fall to floor following a path in the vertical direction. Can you explain using the term "acceleration of gravity" why this the case? How do you understand acceleration of gravity? I already explained how I understand it in post #13.
 
  • #16
The acceleration of gravity is downward and has a magnitude of 9.8 m/s^2. So if they ball falls with no air resistance it will accelerate at that that speed
 
  • #17
mdavies23 said:
The acceleration of gravity is downward and has a magnitude of 9.8 m/s^2. So if they ball falls with no air resistance it will accelerate at that that speed
Can you explain what this means in terms of velocity change? Fill in the blank

For every second that goes by _________________________________________
 
  • #18
kuruman said:
Can you explain what this means in terms of velocity change? Fill in the blank

For every second that goes by _________________________________________
The velocity changes by 9.8 m/s
 
  • #19
mdavies23 said:
The acceleration of gravity is downward and has a magnitude of 9.8 m/s^2. So if they ball falls with no air resistance it will accelerate at that that speed

The 9.8 figure applies on earth, but the current problem refers to Mars (or the Moon?), where the acceleration due to gravity is not known. Finding that acceleration is part of the assigned problem.
 
  • #20
More specifically, for every second you add 9.8 m/s (or g on Mars) in the downward direction to the vertical component of the velocity. So let's go back to my question in post #13. How does vx change in time? You said "it would decreasing in a straight line" Do you still think that's the case? If not, what is the case?
 
  • #21
kuruman said:
More specifically, for every second you add 9.8 m/s (or g on Mars) in the downward direction to the vertical component of the velocity. So let's go back to my question in post #13. How does vx change in time? You said "it would decreasing in a straight line" Do you still think that's the case? If not, what is the case?
Vx would be constant
 
  • #22
Great. Now we can proceed to find the velocity vector. You need two components. The horizontal component does not change in time. Read the statement of the problem very carefully.
Can you tell what the horizontal component is?
What about the vertical component? What do you know about that?
 
  • #23
kuruman said:
Great. Now we can proceed to find the velocity vector. You need two components. The horizontal component does not change in time. Read the statement of the problem very carefully.
Can you tell what the horizontal component is?
What about the vertical component? What do you know about that?
So the horizontal component would stay at 2 m/s and the vertical component will change with g?
 
  • #24
mdavies23 said:
So the horizontal component would stay at 2 m/s and the vertical component will change with g?
Correct. You can now answer part (b) first and find what g is on Mars. Read the statement of the problem keeping in mind that for every second that goes by you must add gMars m/s in the downward (negative) direction.
 
  • #25
kuruman said:
Correct. You can now answer part (b) first and find what g is on Mars. Read the statement of the problem keeping in mind that for every second that goes by you must add gMars m/s in the downward (negative) direction.
So would i have to find the magnitude and direction of the vector given first?
 
  • #26
mdavies23 said:
So would i have to find the magnitude and direction of the vector given first?
How would that help you find the acceleration?
kuruman said:
Read the statement of the problem keeping in mind that for every second that goes by you must add gMars m/s in the downward (negative) direction.
Can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?
 
  • #27
kuruman said:
How would that help you find the acceleration?

Can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?
Oh
kuruman said:
How would that help you find the acceleration?

Can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?
it would be the the slope
 
  • #28
mdavies23 said:
t would be the the slope
The slope of what? The question I asked was "Can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?". You answered "Oh". That's not a very informative answer, so I will ask it a bit differently.
How can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?
 
  • #29
kuruman said:
The slope of what? The question I asked was "Can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?". You answered "Oh". That's not a very informative answer, so I will ask it a bit differently.
How can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?
Wouldn't that be dv/dt to go from velocity to acceleration
 
  • #30
mdavies23 said:
Wouldn't that be dv/dt to go from velocity to acceleration
Actually, this is a 2d situation, so what are dvx /dt and dvy /dt in this case? Can you figure out the numbers from what is given?
 

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