Three particles A, B, and C are thrown from t | vedclass

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(A) According to the law of conservation of energy, the total mechanical energy of a particle remains constant if only gravity acts on it.Let the heig

Vedclass · Accepted Answer

$V_A = V_B = V_C$

Vedclass · Answer

(A) According to the law of conservation of energy, the total mechanical energy of a particle remains constant if only gravity acts on it.Let the height of the tower be $h$ and the initial speed of each particle be $u$.The initial energy of each particle at the top of the tower is $E_i = \frac{1}{2}mu^2 + mgh$.The final energy of each particle just before hitting the ground is $E_f = \frac{1}{2}mV^2 + 0$, where $V$ is the final speed.Since $E_i = E_f$, we have:$\frac{1}{2}mu^2 + mgh = \frac{1}{2}mV^2$$u^2 + 2gh = V^2$$V = \sqrt{u^2 + 2gh}$Since $u$ and $h$ are the same for all three particles, the final speed $V$ will be the same for all three particles regardless of the direction of projection.Therefore, $V_A = V_B = V_C$.

Three particles $A$, $B$, and $C$ are thrown from the top of a tower with the same speed. $A$ is thrown up, $B$ is thrown down, and $C$ is thrown horizontally. They hit the ground with speeds $V_A$, $V_B$, and $V_C$ respectively. Which of the following is correct?

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