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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 must leave the ground to perform this feat.

Answer :

Eduard22slygo

The minimum speed with which the athlete must leave the ground is 2.21 m/s

From the question given above, the following data were obtained:

  • Maximum Height (h) = 0.25 m
  • Final velocity (v) = 0 (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Initial velocity (u) =?

v² = u² – 2gh (going against gravity)

0² = u² – (2 × 9.8 × 0.25)

0 = u² – 4.9

Collect like terms

u² = 0 + 4.9

u² = 4.9

Take the square root of both side

u = √4.9

u = 2.21 m/s

Thus, the athlete must leave the ground with a minimum speed of 2.21 m/s.

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