What Is the Acceleration of a Thrown Ball at Its Peak?

[trollcast]

Homework Statement

Problem A:

A ball is thrown upwards and returns to the throwers hand.
Air resistance is ignored and the upwards direction is taken as positive.
Circle the correct values for the acceleration of the ball:

Going up : g / 0 / -g
At max height: g / 0 / -g
Going down: g / 0 / -g


Problem B:

2 model cars are at opposite ends of a 2000mm track.
When timing starts A is moving with a uniform velocity of 50mms^-1 towards B.
B starts from rest when timing starts, then accelerates uniformly at 4mms^-2 for 10s and then continues towards A with the uniform velocity it has reached.

Calculate the distance from the initial position of A the two cars meet.

Homework Equations

Equations of Motion

The Attempt at a Solution

Problem A:

On the way up the ball is moving in the opposite direction to the force of gravity so g will be -ve as the ball is deccelerating

Then on the way down the opposite is true so g will be +ve.

However at the max height I know the acceleration cannot be 0 as the ball is still being pulled by the force of gravity however I can't decide whether it would be +ve or -ve g.

I think it must be -ve as at the max height it then immediately starts moving downwards so it must have just experienced an acceleration to make it move downwards?

Or should that all be reversed because of the upwards is positive clause?

Problem B:

I calculated the distance traveled by both A and B during the first 10 seconds and got:

A travels 500mm
B travels 200mm

Then there is 1300mm of track left.

A is traveling at 50mms^-1 as it is traveling at constant uniform velocity.
I then calculated that B is traveling at 40mms^-1 in the opposite direction.

Then I looked at it from the point of view of A.

So A is traveling effectively at 90mms^-1 over 1300mm to meet with B.

This then allows me to calculate that that time taken to cover this distance would be 24.444... seconds.

Then I add on the 10 seconds I used earlier.

And use s = 0.5 * (u + v) * t with the values for A to find a distance of 1222.222... mm away from A's starting point?

 

[jedishrfu]

for problem A:

Ask your self if g represents gravitational acceleration has it changed at all after throwing the ball and then catching it?

 

[trollcast]

for problem A:

Ask your self if g represents gravitational acceleration has it changed at all after throwing the ball and then catching it?

So then the acceleration would be -g in all cases as the acceleration vector would be downwards in all cases and its been specified that upwards is positive?

 

[jedishrfu]

So then the acceleration would be -g in all cases as the acceleration vector would be downwards in all cases and its been specified that upwards is positive?

that would be a yes

 

[Nickie]

For Problem A, you are correct in your reasoning. At the max height, the acceleration of the ball would be -9.8m/s^2 (assuming standard gravitational acceleration on Earth). This is because at the max height, the ball has reached its peak and is about to start moving downwards, so it is experiencing a negative acceleration due to gravity.

As for the rest of the problem, your calculations and reasoning seem correct. You are correct in using the equation s = 0.5 * (u + v) * t to find the distance from A's initial position. However, you may want to double check your calculation for the time taken to cover the remaining distance. Overall, your approach to solving the problem is correct and shows a good understanding of the equations of motion.