How Is Kinetic Energy Affected by Different Fall Speeds?

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SUMMARY

This discussion focuses on the effects of fall speeds on kinetic energy and gravitational acceleration. When two objects of the same mass are dropped from a height, one landing 30% faster results in a kinetic energy increase of 69%, calculated using the formula E = 1/2 m v^2. A projectile fired vertically at 30 m/s reaches a height of 44.1 m, losing 2% of its initial energy to air resistance. Additionally, the discussion addresses gravitational acceleration at varying heights and the mass of Earth required to achieve specific gravitational effects.

PREREQUISITES
  • Understanding of kinetic energy formula (E = 1/2 m v^2)
  • Basic principles of gravitational acceleration (g = GM/r^2)
  • Concept of energy loss due to air resistance
  • Knowledge of projectile motion and maximum height calculations
NEXT STEPS
  • Study the relationship between velocity and kinetic energy in depth
  • Learn about air resistance and its impact on projectile motion
  • Explore gravitational force variations with height using GM/r^2
  • Investigate the implications of mass and acceleration on planetary bodies
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Students and enthusiasts of physics, particularly those interested in mechanics, energy calculations, and gravitational effects.

rexhi
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I would really appreciate to know how some problems are solved!

1)Two different objects are dropped from rest off of a 50-m-tall cliff. One lands going 30% faster than the other. The two ojects have the same mass. How much more kinetic energy does one object have at the landing than the other? The answer is 69% more why?

2) A projectile is fired vertically at a speed of 30 m/s. It reaches a maximum height of 44.1 m. What fraction of its initial energy has been lost to air resistance at this point? the answer is 2%, can somebody explain me why?

3) At what height above the Earth's surface is the acceleration due to gravity 10% of that at sea level? 1.36 x 10^7 m

4) The moon has 1/2 the acceleration of gravity of the earth. What would the mass of the Earth have to be to have this acceleration at its surface? 1.03x10^3

I wouuld really appreciate if someone can help me, physics is my weakness :(
 
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1.
E = 1/2 m v^2
This depends on v^2 and so, the ratio of energies will be the ratio of the velocities squared, so E_1/E_2 = (1.3v_1)^2/v_2
which gives 1.69, ie 69% more kinetic energy

2. with no air resitance v^2 = 2gs where v is the initial velocity, g the acceleration due to gravity and s the maximum height. calculation this gives 45, now just go 44.1/45 to give 0.02, ie 2%

3. g = GM/r, use this and the fact that 10% of g is 0.1g, divide g by 0.1 g to get the answer.

4. I am not sure you have this answer right, because that is a migthy small mass for a planet to have. Solve this one by using the equation in 3, with 0.5g at the front.
 
thank you so much for your help SporadicSmile :)

by the way 1.3 was it achieved by (1/2)/(1/2) + 0.3?
 
Actually,

<br /> g = GM/r^2 <br />
 

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