Power of throwing a ball up.

[rainman123]

What is the power neccesary to throw a ball vertically up against gravity? Lets say we want to know how much power a human spends on throwing a ball up.

 

Lets say we know the work done W1 (from the bottom to up phase of the ball), and we also know the work done W2 (from top to bottom of the ball). Lets say that we also know the time it takes for the ball to go up T1, and also the time it takes for it to go down T2. The two individual terms should be equal in both up and down phases respectively for the conserved system.

 

But since this system is conserved and in a closed loop, is the total work of the system zero (W_system=W1+W2=0)? But then Power spent shouldnt be zero.

 

My question is, what is the power used by the human throwing the ball up?

 

Is the actual power used to throw the ball up calculated as Power_human=W1/T1, only? So just the power for the up phase. The Power P=W2/T2 is the power of the gravity right? That is such that the total work done by the system in this closed loop is zero. Also it is wrong to say that the power of the human is Power=W1/(T1+T2).

[swansont]

You've done work in getting the ball to leave your hand at speed v (1/2 mv^2), and also raising its potential energy by mgh where h is the distance between where you start to throw the ball and its release point. The average power is going to be that amount of work divided by the time it took you to throw the ball.

[rainman123]

hi, ok so you are saying then there are esentially 4 separate events.

 

(1) first the ball is zero velocity on my hand.

 

(2) second event is the ball accelerates to V as it leaves my hand (assuming potential energy gain here is small)

 

(3) third event is the ball reaches a maximum height and zero kinetic energy with max potential energy.

 

(4) Fourth event is the ball falls back to my hand by losing potential and gaining kinetic energy.

 

So you are saying to calculate the power used by the human to throw this ball, then it is the work it took from event 1 to 2 divided by the time between event 1 and 2. Power_human=W_12/T_12.

 

Can we not calculate Power_human by W_23/T_ 23? Not possible because W_23 = mgh-1/2mv^2, and since energy is conserved so W_23=0?

 

Finally another question is, it will be totally wrong to suggest Power_human=W_12/(T_12+T_23+T_34)

[swansont]

hi, ok so you are saying then there are esentially 4 separate events.

 

(1) first the ball is zero velocity on my hand.

 

(2) second event is the ball accelerates to V as it leaves my hand (assuming potential energy gain here is small)

 

(3) third event is the ball reaches a maximum height and zero kinetic energy with max potential energy.

 

(4) Fourth event is the ball falls back to my hand by losing potential and gaining kinetic energy.

 

So you are saying to calculate the power used by the human to throw this ball, then it is the work it took from event 1 to 2 divided by the time between event 1 and 2. Power_human=W_12/T_12.

 

 

Yes.

 

There is no work being done on the ball after it leaves your hand, which is why the sum of KE + PE, which is the work you originally did, remains constant.

[rainman123]

ya ok. But is there a way to figure out the power consumed by the human, if the only information we know is the 2-4th event.

 

THat is, we only know the velocity leaving the hand, the height it gains when it reaches the top at v=0, and the time it takes to get up there. Is it possible then to figure out the power consumed by the human?

 

Is the answer no because we are not told the throwing event by the human, that is we don't know the time it took the human to throw the ball up to the velocity V. Althought we do know that energy is conserved, that is E=KE+PE.

 

In other words, if we only know the velocity V as it leaves the hand, the time T it takes to reach maximum height and the maximum height of the ball, then there is no way to figure out the power consumed by the human right?

[Mokele]

Well, no matter what else, you need to know the mass of the ball in order to find Watts.

 

However, you could make assumptions about human arm length, which could get you what you need.

[swansont]

In other words, if we only know the velocity V as it leaves the hand, the time T it takes to reach maximum height and the maximum height of the ball, then there is no way to figure out the power consumed by the human right?

 

The T and V you give are not independent — if you know one, you can find the other. They tell you the energy of the ball as it is tossed (assuming you know the mass, as Mokele has pointed out), but without knowing how long it took, you can't calculate the power.

[rainman123]

Yea I know V and T given we know the decelleration constant, we can find V and T as a function of distance.

 

Lets say we know the mass.

 

Ok so it is true then that there is no way to determine the power consumed by the human throwing the ball if the only pieces of information are given:

 

(1) Mass of Ball

(2) Velocity of Ball leaving hand

(3) Max Height Gained

 

This is because the power used by the human can only be determined given we know the parameters during the throwing of the ball. So in order to know the power consumed by the human we need to know the time it took the human to throw the ball from zero velocity to the maximum velocity as it leaves the hand. In this case Power=KE_human/T_throwing, the kinetic energy KE_human we can determine, but the time of throwing T_throwing we dont know.

 

In other words, my original question is indeterminable given the information provided since we know nothing about the throwing incident.