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Three particles A, B and C are projected from the top of a tower with the same speed. A is thrown straight upwards, B straight down and C horizontally. They hit the ground with speeds $ν_{A}$ , $ν_{B}$ and $ν_{C}$ then which of the following is correct?

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$v_{1} = v_{B} = v_{C}$

$v_{1} > v_{1} > v_{1}$

$v_{A} = v_{B} > v_{C}$

$v_{1} > l_{13} = v_{0}$

answer is B.

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Detailed Solution

Change in potential energy for all three particles is same. Hence, change in kinetic energy will also be same.

or $v_{A} = v_{B} = v_{C}$

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