A body of mass 2 kg thrown up vertically with K.E of 490 joules.if the acceleration due to gravity is 9.8 m/s^2,then the height at which the K.E of the body becomes half its orginal value is given by
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hii mate here is the answer to your question: -
given-- mass (m)=2kg.
K.E=490 joule.
acceleration due to gravity (g)= 9.8 m/s^2.
so we'll have to find the 'velocity' first;
therefore, 1/2mv^2=490----(1)
implies, 1/2×2×v^2=490
---->>v^2=490----(2)
therefore, the height at which the kinetic energy is half of its original value will be 4900/196/2=12.5
hope this helps you
keep smiling
given-- mass (m)=2kg.
K.E=490 joule.
acceleration due to gravity (g)= 9.8 m/s^2.
so we'll have to find the 'velocity' first;
therefore, 1/2mv^2=490----(1)
implies, 1/2×2×v^2=490
---->>v^2=490----(2)
therefore, the height at which the kinetic energy is half of its original value will be 4900/196/2=12.5
hope this helps you
keep smiling
prefectone:
sorry
4900/196
=25
so at height =25m kinetic energy is 490 j
so at height=25/2 =12.5m k.e is halved that means 245j
k tqs
by the way what is ur name
??
stylee
where do u live?
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