Question

Three balls of equal mass are fired simultaneously with equal speeds from the same height above ground. Ball 1 is fired straight up, ball 2 is fired straight down, and ball 3 is fired horizontally. Rank in order, from largest to smallest, their speeds v1, v2 and v3 as they hit the ground. Explain using the Kinetic Energy / Work Theorem.

Answer 1

Answer:

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Answer 2

The order, from largest to smallest, of speeds of balls as they hit the ground is:  V1 = V2 = V3

• This is based on the conservation of energy. We know that the mechanical energy is equal to the sum of kinetic and gravitational energies and is constant at each point.
• Applying E=K+U at the initial and final point for each of the three balls, we get:

At the initial point,

$E=\frac{1}{2} m u^{2} + mgh$ (this is same for all the balls since all are initially thrown with the same velocity from the same height and have the same masses.

At the final point,

U=0, thus balls have only kinetic energies

Since masses are same and total energies are same

Hence, all balls have the same final velocities.