How Fast Does a Ball Travel When Dropped from a Height?

AI Thread Summary
A ball with a mass of 0.163 kg is dropped from a height of 1.89 m, and the acceleration due to gravity is 9.8 m/s². To find the speed of the ball at a height of 0.997 m, the gravitational potential energy is converted into kinetic energy. The calculation shows that the velocity at this height is approximately 6.08 m/s, but it is noted that this approach assumes complete energy conversion, which may not be accurate. The correct formula to use for this scenario is Vf = sqrt(2a * delta y). The discussion emphasizes the importance of checking the height and using the appropriate equations for accurate results.
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Homework Statement




A ball of mass 0.163 kg is dropped from a
height 1.89 m above the ground.
The acceleration of gravity is 9.8 m/s^2
Neglecting air resistance, determine the
speed of the ball when it is at a height 0.997 m
above the ground.
Answer in units of m/s

Homework Equations



Gravitational Energy = Kinetic Energy

Mass x Acceleration From Gravity x Height = 1/2 (mass)x(velocity)^2

Acceleration From Gravity x Height = 1/2 (velocity)^2 [mass cancels]



The Attempt at a Solution




(9.8 m/s^2) x (1.89m) = 1/2 (velocity)^2

Velocity^2 = 2[(9.8 m/s^2) x (1.89m)]

Velocity = ~6.08 m/s
 
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Physics-Pure said:
Gravitational Energy = Kinetic Energy

Mass x Acceleration From Gravity x Height = 1/2 (mass)x(velocity)^2

This is correct only when one kind of energy is converted entirely into another. Which is not the case in this problem. What equation should you really use?
 
Vf = sqrt 2a delta y?
 
Physics-Pure said:

Homework Statement




A ball of mass 0.163 kg is dropped from a
height 1.89 m above the ground.
The acceleration of gravity is 9.8 m/s^2
Neglecting air resistance, determine the
speed of the ball when it is at a height 0.997 m
above the ground.
Answer in units of m/s
[]

Check the height.
 
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