Magnitude and direction of impulse

[PhunWithPhysics]

I have a ball that is dropped from from rest at a height of 1.2 meters it hits the ground and bounces to a height of 0.7 meters. The mass of the ball is 0.5kg. I am assuming that initial velocity is 0 because it is released from rest. I need to find the magnitude of direction of the impulse of the net force.
I know that the direction will be in the upward direction. But how do I find Impulse. I have Impulse (J)=average forse X the change in time.
net force= (MVfinal-MVinitial)/change in time. ? Were can I go from here?
Any help would be great thanks.

[Tide]

The impulse is the change in momentum.

[PhunWithPhysics]

how can I find impulse (J) which equals average force X change in time. when I need velocity to find the change in momentum?

[Doc Al]

You certainly do need the velocities to find the change in momentum. So figure them out! You have the distance that the object fell from and the distance it rose to. That's enough information to figure out how fast it was moving just before and just after the collision with the ground.

[PhunWithPhysics]

How can I find velocity if all I have is hieght? Don't I need time to find velocity? I know that the answer is 4.28 , I just don't under stand what equation I need to use? still confused.

[Doc Al]

You need to be able to solve this problem: When an object is dropped from a given height, how fast is it moving when it hits the ground?

Hint: You can use kinematics of falling objects to find the speed, or you can use energy methods.

[daniel_i_l]

Remember that the total mechanical energy is constant in each half of the trip (it is constant on the way down and on the way up, but not during the whole trip cause some energy goes into the ground)

[PhunWithPhysics]

If i use kinematics of a falling object they all use time or velocity and energy measurements use velocity too don't they? I need help.

[Doc Al]

Given distance and acceleration, one can certainly find the final speed using kinematics. Hint: If you don't see how to do it in one step, do it in two steps. Find the time, then use the time to find the speed.

Similarly, using conservation of energy (realizing that the decrease in PE equals the increase in KE) one can find the speed given the distance fallen. Try it and see.

[PhunWithPhysics]

I was not given the acceleration though. Just Height final, height initial and mass?

[Tide]

You need to be able to solve this problem: When an object is dropped from a given height, how fast is it moving when it hits the ground?
Hint: You can use kinematics of falling objects to find the speed, or you can use energy methods.

Actually, it's the dynamics of falling objects that you are interested in and not kinematics. The latter refers to motion without references to mass and force such as "velocity is time rate of change of position" etc.

[Tide]

I was not given the acceleration though. Just Height final, height initial and mass?

You're dealing with acceleration due to gravity and you would not have been assigned this problem unless you had already covered it in your course. Sometimes you actually get to use stuff you have already learned in prior lessons. The number you are looking for is g = 9.8 m/s^2.

[PhunWithPhysics]

I had already been toying with the acceleration as 9.8 which is m/s^2.
So that times the mass will give me (Kg) times m/s^2 which equals Newtons then I just need to find time. Or am I going about this all wrong. Can you explain what equation I should be using. Or how to re-arrange a certain equation to get the right one. I've tried everything from putting it into Potential energy equations to Impulse equations to Force net equations. I'm just not quite understanding it? Any more help you can give would be great.

[Tide]

If you drop an object initially at rest from some height h its speed is v = \sqrt {2 g h} at the end of its fall. Does that ring any bells?

[PhunWithPhysics]

Is h = h,final - h,intial, or the other way, V = the square root of 2 times 9.8 times h , but what can I do with velocity to get the impulse to be 4.28 because I keep coming up with 3.7 something what am I doing wrong. What else am I missing. I know the answer is 4.28 but how did it get to that?

[Tide]

What you did gives you the speed just before impact with the floor. You have additional information regarding how high the ball bounces. Using the same kind of reasoning you can determine from this how fast the ball was travelling immediately after the bounce. You need both of those speeds to calculate the impluse or change in momentum during the bounce.

[daniel_i_l]

In the equation that tide wrote, if h is the initial hight then v would be the speed right before hitting the ground, if h is the final hight then v is the speed right after the ball leaves the ground. With the final and initial speeds, you can find the final and initial momentum that will give you the impulse.

[PhunWithPhysics]

I can't figure it out so just don't worry about it any more