Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speed V_{A ,}V_B,and V_cthen compare the three velocities.

V_{A} = V_{B} = V_{C}

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V_{B} > V_{C} > V_{A}

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V_{A} = V_{B} > V_{C}

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V_{A} > V_{B} = V_{C}

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Solution

The correct option is (A) $V_{A} = V_{B} = V_{C}$
As all three of them have same potential energy due to height.
$w o r k e n e r g y t h e o r e m$
$W_{a l l} = Δ K E$
$W_{P E d u e t o h e i g h t} = Δ K E$
$m g h = \frac{m v^{2}}{2}$
use this logic for all three cases
by using work energy theorem
$∴ V_{A} = V_{B} = V_{C}$

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Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speed VA ,VB ,and Vc then compare the three velocities.

The correct option is (A) VA=VB=VCAs all three of them have same potential energy due to height. workenergytheoremWall=ΔKEWPEduetoheight=ΔKEmgh=mv22use this logic for all three casesby using work energy theorem ∴VA=VB=VC

Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speed V_{A ,}V_B,and V_cthen compare the three velocities.