Question

A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 j. the height at which the kinetic energy of the body becomes half of the original value is ______.

Answer 1

Let u be the velocity with which it was projected
K = 0.5mu²
490 J = 0.5 × 5 kg × u²
u² = 196 J/kg

Let v be the velocity at the point where it’s kinetic energy becomes half is
490/2 = 0.5 mv²
490 Joule = 5 kg × v²
v² = 98 Joule/kg

Now, use equation of motion v² - u² = 2aS
h = [(98 - 196) J/kg] / (2 × -9.8 m/s²) = 5 m

K.E will become half at height of 5 m

Answer 2

Answer:

K.e+ p.e=0

Put 1/2 mv2 and mgh in k.e nd p.e respectively.

Take m common in both

1/2 mv2 = -mgh

m canceled out.

1/2 v2= gh

or h=v2/2g put values accordingly n ÷ by 2.

Point out if any errors.

Shortcut

∆h=K.E/2mg.

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