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- Apr 16, 2010
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I'm have trouble trying to figure out what I'm doing wrong here:
The question states:
A 2 kg ball is thrown upwards with a speed of 40 m/s. At what height will the ball be, when its kinetic energy is equal to its potential energy?
I know that during half the trip (half the height), Potential Energy = Kinetic Energy. In order to solve for height, I use one of the kinematic equations.
But here's where the problem lies, I get two different answers using different equations but can't seem to understand why. They should be equal regardless.
h = hi + vit + .5at^2
h = 0m + (40m/s)(4s) + (5 m/s)(4^2)
h = 240 meters
or using this equation:
vf^2 = vi^2 + 2ah and solving for 2
2gh = vi^2
2(10m/s)h =1600 (m/s)^2
h = 1600 (m/s)^2 / 20
h =80 meters
Any idea what I'm doing wrong to get different answers?
The question states:
A 2 kg ball is thrown upwards with a speed of 40 m/s. At what height will the ball be, when its kinetic energy is equal to its potential energy?
I know that during half the trip (half the height), Potential Energy = Kinetic Energy. In order to solve for height, I use one of the kinematic equations.
But here's where the problem lies, I get two different answers using different equations but can't seem to understand why. They should be equal regardless.
h = hi + vit + .5at^2
h = 0m + (40m/s)(4s) + (5 m/s)(4^2)
h = 240 meters
or using this equation:
vf^2 = vi^2 + 2ah and solving for 2
2gh = vi^2
2(10m/s)h =1600 (m/s)^2
h = 1600 (m/s)^2 / 20
h =80 meters
Any idea what I'm doing wrong to get different answers?