Physics in vertical jump and basball hit

AI Thread Summary
To calculate the force of a vertical jump reaching 6 feet in 0.797 seconds, one must consider the mass center and the initial velocity at take-off, which is often underestimated. The discussion highlights the need for realistic assumptions about force variation during the jump, as constant power may be more accurate than constant force. For the baseball scenario, knowing the mass of the ball (5 oz) and the coefficient of restitution is crucial for determining bat speed, which is not directly affected by bat diameter. Additionally, calculating the distance a runner would slide on ice after running at 15 mph requires more data, such as friction coefficients. Overall, understanding the physics principles involved is essential for accurate calculations in these scenarios.
Michael12438
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Homework Statement


Okay so these are from movies. I am trying to prove them right or wrong using physics. how can I figure out the pounds of force of a vertical jump when I have the:

height of the jump = 6 feet
Time to reach top = .797 sec
Mass of person = 100lbs
2. How can I figure out this problem...

If a baseball is thrown with a velocity of 90mph, and the ball exits the bat at 140mph, what eas the bat speed? Consider the ball was hit with a wood bat. On its most dense part with the barell diameter of 2.5 in, the bat is 30 oz

Friction- how far would a runner slide if he fell on ice after running 15 mph

Homework Equations

The Attempt at a Solution


For the first one I messed up somehow using the fact that initial velocity was 0m/s^2 and I got that the force would be only around 50lbs. The number should be in te thousands I'm suspecting. Please help
 
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Michael12438 said:
the pounds of force of a vertical jump
You have to make some assumption about how the force varies over the period before leaving the ground. Constant is a bit unrealistic as it means power increases linearly with speed. Constant power might be nearer the mark.
Michael12438 said:
height of the jump = 6 feet
Be careful how you define that. It's the movement of the mass centre that's important for the forces etc. Most vertebrates can only get their mass centres up about three feet from the point of leaving the ground.
Michael12438 said:
Time to reach top = .797 sec
From starting the jump (from crouching? How low?) or from lift off?
Michael12438 said:
If a baseball is thrown with a velocity of 90mph, and the ball exits the bat at 140mph
You will also need to know the mass of the ball; and the coefficient of restitution... or you could just suppose it to be perfectly elastic.
Michael12438 said:
On its most dense part with the barell diameter of 2.5 in
I don't see how the diameter matters. I guess you mean the bat is struck at its mass centre.
Michael12438 said:
Friction- how far would a runner slide if he fell on ice after running 15 mph
Again, you need more data. What do you think that might be?
 
For the friction what other data do I need?

The mass of the ball is 5 oz

The height of the jump is 6 feet. It is from a movie and I'm trying to prove it is not realistic

I don't know what restitution is.
 
The jump time is from lift off and using the mass center
 
Michael12438 said:
For the friction what other data do I need?
You are trying to calculate a distance given an initial speed and a final speed. You don't know the time, so what else will help?
Michael12438 said:
The mass of the ball is 5 oz
What equations do you know related to collisions? What is or may be conserved?
Michael12438 said:
I don't know what restitution is.
It's a measure of how well bodies rebound from collisions. Take two bodies, initially moving at speeds u1, v1. Positive is to the right, say. Take it all to be in one straight line. After collision, they move off with speeds u2, v2. It turns out that if you were to repeat the experiment with the same bodies at different speeds you would find there is a constant (usually written as e, for elasticity, or R for restitution) such that v2-u2 = R(u1-v1). (Note the sign reversal.) For more info G**gle for coefficient of restitution.
In a perfectly elastic collision, the coefficient is 1. The fraction of energy retained is the square of the coefficient.
Michael12438 said:
The jump time is from lift off and using the mass center
Mass centre up 6 feet - definitely unrealistic.
What take-off speed would be needed? What SUVAT equations do you know? (You should have listed these in the OP.)
 
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