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Question

A particle is thrown upwards with speed v0. Find its speed when it reaches 2/3rd of the maximum height.

A
v02
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B
2v03
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C
v03
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D
v03
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Solution

The correct option is C v03
Maximum height of the particle is given as
Hmax=u22g=v202g
So, the speed of the particle at the height (23×Hmax=v203g) is given as
v2u2=2as
v2v20=2×(g)×v203g
v2=v202v203=v203
v=v03

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