Question #18106

an olympic high jumper leaps over a horizontal bar. the jumper's center of mass is raised 0.25 m during the jump. calculate the minimum speed with which the athlete must leave the ground to perform this feat.

Expert's answer

From the energy conservation

(min. kinetic energy at start) = (potential energy gain at max. height)

m*v^2 / 2 = m*g*h;

v = sqrt(2*g*h) = sqrt(2 * 9.8 m/s^2 * .25 m) = 2.214 m/s

Answer: 2.214 m/s

(min. kinetic energy at start) = (potential energy gain at max. height)

m*v^2 / 2 = m*g*h;

v = sqrt(2*g*h) = sqrt(2 * 9.8 m/s^2 * .25 m) = 2.214 m/s

Answer: 2.214 m/s

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