What is sign of y if ball is thrown up?

[freshyy]

This is a general question that i am confused about. If I throw a ball up, then is the distance that it reaches at the very top the y or y-initial? Also, is this y or y-initial positive or negative?

 

[berkeman]

This is a general question that i am confused about. If I throw a ball up, then is the distance that it reaches at the very top the y or y-initial? Also, is this y or y-initial positive or negative?

You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

Does that help?

 

[freshyy]

You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

Does that help?

Yes, thank you

 

[freshyy]

You can define your coordinates however is the most convenient for solving any particular problem. The simplest 2-d coordinate system for projectile motion problems would be to put the origin on the ground, with the x-axis horizontal and the y-axis vertical. That means y=0 is at ground level, so if you are 6 feet tall and throw a ball up, then y(t=0)=6', and Vy(0)= whatever velocity you release the ball at time t=0. The height at the top of the ball's travel will depend on Vy(0) at the release point y(0)=6' plus however long it takes for the ball to stop moving up. Use an equation like this...

y(t) = y(0) + Vy(0)*t + 1/2*ay*t^2

Where a = the acceleration due to gravity, which in this case is -9.8m/s^2.

Does that help?

Also, the time that the ball takes to come down is twice the time the ball takes to reach its maximum height, right?

 

[berkeman]

Also, the time that the ball takes to come down is twice the time the ball takes to reach its maximum height, right?

Not if you release it from your 6' height. Right?

 

[berkeman]

Well, I mean the two times from release to apex, and from apex to hitting the ground. Those are not equal.

 

[freshyy]

Well, I mean the two times from release to apex, and from apex to hitting the ground. Those are not equal.

Yea, in understand it now thank you.