Question

Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the 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 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

Answer 1

Answer:

d) v1 = v2 = v3

Explanation:

This can be answered using conservation of energy. We calculate the mechanical energy E=K+U (sum of kinetic and gravitational potential energies) at the original and final points, and impose they are equal.

At the original point we have, for the three balls:

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

At the final point we have, for the three balls:

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

Since we have $E_i=E_f$, and $E_i$ is the same for all balls, then $E_f$ is the same for all balls, which means that $v_f$, the final velocity, is the same for all balls.