Calculating the velocity and acceleration given vector velocity at 1 second

[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

 

[kuruman]

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".

 

[haruspex]

If the gravitational acceleration is g_M, how does the vertical velcity vary with time?

I presume the reference to moon in part b is a mistake.

 

[mdavies23]

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?

 

[kuruman]

What does maximum height mean?

 

[mdavies23]

What does maximum height mean?

velocity equals 0

 

[kuruman]

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.

 

[mdavies23]

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

 

[kuruman]

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.

 

[mdavies23]

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

 

[kuruman]

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?

 

[mdavies23]

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.

 

[kuruman]

I would think the Vy would be zero?

That is correct, v_y is instantaneously zero. It starts at v_0y then is reduced to zero because every second that goes by, g m/s are added in the "down" direction to the v_y that is already there. That's what the equation v_y = v_0y - gt says in the language of mathematics. What about v_x? How does that change as time goes by?

 

[mdavies23]

It would be a decreasing straight line

 

[kuruman]

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.

 

[mdavies23]

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

 

[kuruman]

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 _________________________________________

 

[mdavies23]

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

 

[Ray Vickson]

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.

 

[kuruman]

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 v_x 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?

 

[mdavies23]

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 v_x 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

 

[kuruman]

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?

 

[mdavies23]

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?

 

[kuruman]

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 g_Mars m/s in the downward (negative) direction.

 

[mdavies23]

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 g_Mars m/s in the downward (negative) direction.

So would i have to find the magnitude and direction of the vector given first?

 

[kuruman]

So would i have to find the magnitude and direction of the vector given first?

How would that help you find the acceleration?

Read the statement of the problem keeping in mind that for every second that goes by you must add g_Mars m/s in the downward (negative) direction.

Can you deduce the acceleration if you know the velocity at two time intervals 1 second apart?

 

[mdavies23]

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

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

 

[kuruman]

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?

 

[mdavies23]

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

 

[kuruman]

Wouldn't that be dv/dt to go from velocity to acceleration

Actually, this is a 2d situation, so what are dv_x /dt and dv_y /dt in this case? Can you figure out the numbers from what is given?

 

[mdavies23]

Actually, this is a 2d situation, so what are dv_x /dt and dv_y /dt in this case? Can you figure out the numbers from what is given?

dvx/dt would be 0

 

[kuruman]

dvx/dt would be 0

Yes, it would be. What about dv_y /dt? You know it's constant and equal to g_Mars. How can you find a value for it? What do you need to know?

 

[mdavies23]

Yes, it would be. What about dv_y /dt? You know it's constant and equal to g_Mars. How can you find a value for it? What do you need to know?

The velocity at time t=2

 

[kuruman]

The velocity at time t=2

You need to know more than that. Remember what we said, "For every second that goes by, you add g_Mars m/s in the downward direction to whatever velocity is already there." Acceleration has to do with change in velocity. You cannot determine a change from just one number. Change is what you end up with minus what you start with. You need two numbers to find the change in v_y. One number is v_y at t = 2s. What could the other number be?

 

[mdavies23]

You need to know more than that. Remember what we said, "For every second that goes by, you add g_Mars m/s in the downward direction to whatever velocity is already there." Acceleration has to do with change in velocity. You cannot determine a change from just one number. Change is what you end up with minus what you start with. You need two numbers to find the change in v_y. One number is v_y at t = 2s. What could the other number be?

V_y at time t=1