Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st

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6 days ago

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Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st

raight up, ball 2 is fired straight down, and ball 3 is fired horizontally. Rank in order from largest to smallest the speeds of the balls, v1, v2, and v3, just before each ball hits the ground. a. v1 > v2 > v3 b. v3 > v2 > v1 c. v2 > v3 > v1 d. v1 = v2 = v3

Physics

1 answer:

Ostrovityanka [942] · Answer 1 · 2026-02-26T23:16:20+03:00

Answer:

d) v1 = v2 = v3

Explanation:

This can be determined through the principle of energy conservation. We assess the total mechanical energy E=K+U (the sum of kinetic energy and gravitational potential energy) at both the initial and final positions, ensuring they remain constant.

<pInitially, for the three spheres, we have:

$E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh$

Finally, for the three spheres, we see:

$E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}$

<pGiven that $E_i=E_f$ , and since $E_i$ remains identical for all spheres, it follows that $E_f$ is identical for all spheres, indicating that $v_f$ , the final velocity, is equal for each ball.