A body is dropped from rest from a height h under gravity. The velocity v just before collision with earth will be given by
Answers
Explanation:
When the body strikes the ground all its potential energy is converted to kinetic energy. Therefore it’s kinetic energy will be:
MgH
It’s velocity can be found from:
1/2*mv^2 =mgH
Hence v = sqrt(2gH)
The velocity can also be calculated from the constant acceleration kinematic formulas.
v =sqrt(2as)
And here a =g and s =H, leading of course to the same result.
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Given : a body is dropped from rest from height h gravity
the velocity v just before Collision with earth
To Find : Velocity V
a)v=u+gt
b)v=½gt²
c)v²=2gh
d)v²=u²+2gh
Solution:
PE at height h = mgh
KE at height h = 0 as at rest
TE = PE + KE = mgh
PE at earth = 0 as h = 0
KE = (1/2)mv²
TE = (1/2)mv²
Equate Total Energy
(1/2)mv² = mgh
=> v² = 2gh
Hence v²=2gh is the correct answer
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