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

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(A) According to the law of conservation of mechanical energy, the total mechanical energy at the top of the tower must be equal to the total mechanic

Vedclass · Accepted Answer

$V_A = V_B = V_C$

Vedclass · Answer

(A) According to the law of conservation of mechanical energy, the total mechanical energy at the top of the tower must be equal to the total mechanical energy at the ground level.Let $m$ be the mass of each particle, $h$ be the height of the tower, and $u$ be the initial speed.At the top: $E_i = mgh + \frac{1}{2}mu^2$At the ground: $E_f = 0 + \frac{1}{2}mV^2$Since $E_i = E_f$, we have $mgh + \frac{1}{2}mu^2 = \frac{1}{2}mV^2$.This simplifies to $V^2 = u^2 + 2gh$, or $V = \sqrt{u^2 + 2gh}$.Since $u, g$, 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 $u$. Particle $A$ is thrown upwards, $B$ is thrown downwards, and $C$ is thrown horizontally. If they hit the ground with speeds $V_A, V_B$, and $V_C$ respectively, then:

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