What Height Does KE Become Half for a 2 kg Body Thrown Upward?

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SUMMARY

The discussion centers on determining the height at which the kinetic energy (KE) of a 2 kg body, initially thrown upward with 490 joules of KE, becomes half its original value. Using the equation for potential energy (PE = mgh) and the relationship between KE and PE, the correct height is calculated as 25 meters. This conclusion is reached by setting the initial KE equal to the PE at height H, leading to the equation 490 = 2 * 9.8 * H, which simplifies to H = 25 m. The confusion regarding whether to halve the KE value is clarified, confirming that the question pertains to the height at which KE is reduced to half.

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Homework Statement


A body of mass 2 kg thrown vertically with KE of 490 joules. If the acceleration due to gravity is 9.8 m/s^2 of the body becomes half its original value is given by:

a.50
b.12.5
c.25 m
d.10 m

Homework Equations


PE=MGH
K.E=1/2MV^2
PE=KE

The Attempt at a Solution



K.E=P.E
490=MGH
490=2*9.8*H
490=19.6*H
490/19.6=H
25=H
H=25 m

is it right ? or we have to half the value.
 
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What is the question? The second sentence is incoherent. My best guess is that it is asking how high the body needs to go to lose half of its kinetic energy due to gravity.

In that case, the Kinetic Energy that is lost and the Potential Energy that is gained are both half of 490 joules.
 
yes sorry the sentence is . "the height at which the K.E of the body becomes half its original value is given by"
 

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