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# velocity-time graph watch

1. Could someone please check these notes and see if there are any mistakes ?
Also I don't get the bit that I have wrote in red. I thought the ball went back to its original place at D and that was where it would be caught ?

As the ball moves higher it decelerates I know that gravity and weight have a part in it but I am really confused :\
Thankyou
2. (Original post by yorobun)

Could someone please check these notes and see if there are any mistakes ?
Also I don't get the bit that I have wrote in red. I thought the ball went back to its original place at D and that was where it would be caught ?

As the ball moves higher it decelerates I know that gravity and weight have a part in it but I am really confused :\
Thankyou
I'm not sure about your graph; for a ball thrown upwards, it should be a straight line from a point above zero on the v-axis, through the t-axis (when v=0, the ball is at its maximum height), to a point at an equal distance below the t-axis as the first point was above it.

Is the ball being released at O or A on your graph? If it is being released at O, your graph is incorrect; once it is only being acted upon by gravity, it cannot accelerate upwards. Similarly, when it is caught, the line should become almost vertical until it reaches 0 m/s.

Think about a ball following your graph in real life. It accelerates towards Earth at a constant rate. This acceleration makes the velocity more and more negative, at a constant rate.

Now for gravity and weight; I will try to make this as simple as possible.

All objects have mass; an object's mass is constant, wherever it is in the universe.

An object's weight changes depending on where it is. Weight is a force due to gravity. If an object is in an imaginary place where there is no gravity (i.e. there is nothing else with mass within an infinite distance), it will have no weight.

The Earth attracts the ball via gravity; the force (weight) is equal to the mass of the ball multiplied by the gravitational acceleration (m*g). This is clear from F=ma, where g is a. The value of g actually changes as you move away from Earth into space, but we assume that it is constant for problems like this, hence the straight line on your velocity-time graph (the gradient is the rate of change of velocity, also known as the acceleration; the acceleration is constant, and so is the gradient).

The gravitational force acting on the ball, mg, is its weight. In this problem, the ball's weight is constant because its mass is constant, and we assume that g, the gravitational acceleration, is constant.

Think about F = ma where a = g. F is the force acting on the ball; this is called its weight.

Sorry if this is confusing; I'll happily reword bits if you need me to.
3. 0 is where the ball is at rest and the line 0-A is when the person gives the an uniform acceleration. (thats what it says on the note that my teacher gave me :\ ) So I think the ball is actually moving upwards from point C ?

Thankyou for the clarification about the gravitational force xD

(Original post by ombtom)
I'm not sure about your graph; for a ball thrown upwards, it should be a straight line from a point above zero on the v-axis, through the t-axis (when v=0, the ball is at its maximum height), to a point at an equal distance below the t-axis as the first point was above it.

Is the ball being released at O or A on your graph? If it is being released at O, your graph is incorrect; once it is only being acted upon by gravity, it cannot accelerate upwards. Similarly, when it is caught, the line should become almost vertical until it reaches 0 m/s.

Think about a ball following your graph in real life. It accelerates towards Earth at a constant rate. This acceleration makes the velocity more and more negative, at a constant rate.

Now for gravity and weight; I will try to make this as simple as possible.

All objects have mass; an object's mass is constant, wherever it is in the universe.

An object's weight changes depending on where it is. Weight is a force due to gravity. If an object is in an imaginary place where there is no gravity (i.e. there is nothing else with mass within an infinite distance), it will have no weight.

The Earth attracts the ball via gravity; the force (weight) is equal to the mass of the ball multiplied by the gravitational acceleration (m*g). This is clear from F=ma, where g is a. The value of g actually changes as you move away from Earth into space, but we assume that it is constant for problems like this, hence the straight line on your velocity-time graph (the gradient is the rate of change of velocity, also known as the acceleration; the acceleration is constant, and so is the gradient).

The gravitational force acting on the ball, mg, is its weight. In this problem, the ball's weight is constant because its mass is constant, and we assume that g, the gravitational acceleration, is constant.

Think about F = ma where a = g. F is the force acting on the ball; this is called its weight.

Sorry if this is confusing; I'll happily reword bits if you need me to.
4. (Original post by yorobun)
0 is where the ball is at rest and the line 0-A is when the person gives the an uniform acceleration. (thats what it says on the note that my teacher gave me :\ ) So I think the ball is actually moving upwards from point C ?

Thankyou for the clarification about the gravitational force xD
The ball is not moving upwards from point C!

In this question we are taking a positive velocity to mean away from Earth, and a negative velocity towards Earth.

Therefore, when the line is above the t-axis, the ball is moving upwards. So as the person accelerates it upwards, the velocity is positive; once it is released from the person's hand, it is still moving upwards, but at a smaller and smaller speed. It reaches its maximum height at B, when the velocity reaches 0, and then starts moving back towards Earth – a negative velocity; the line is below the t-axis. It is below the t-axis from B to D, so we know that it does not move upwards during this time.

From the point C to D, the ball is still moving downwards, towards Earth, but with a smaller and smaller speed.

Think about it happening in front of you; if the ball has a large negative velocity at C, it is travelling downwards quickly. It cannot suddenly be travelling upwards; it has to reduce its velocity from a negative value to 0 before it can go upwards again. It's similar to temperature; an object can't immediately go from -100 to 100 degrees – it has to gradually increase its temperature.

Hope this helps.
5. Sorry that was so stupid and I get it cause from C the ball would still be moving downwards but at greater acceleration since the gradient shows the acceleration , right ? And at D it is stationary unless a force is applied to move it upwards or downwards xD
Thankyou so much ^-^
6. (Original post by yorobun)
Sorry that was so stupid and I get it cause from C the ball would still be moving downwards but at greater acceleration since the gradient shows the acceleration , right ? And at D it is stationary unless a force is applied to move it upwards or downwards xD
Thankyou so much ^-^
Yep; the hand catching the ball exerts a force upwards, so the acceleration goes from 9.8 (=g) downwards to a large value upwards, hence the steeper gradient.

Make sure you quote people when you reply to them!

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