Question

In a vertical jump test, the subject's center of mass starts at 1.2 m above the ground at the instant before takeoff. Their center of mass reaches 3.0 m above the ground at the peak of her vertical jump. Given this information, which equation would you need to use to directly calculate the subject's initial velocity at takeoff during the vertical jump test?

vf = vi + at
vf^2= vi^2+ 2a(rf - ri)
v = Δr / Δt
rf - ri = vit + 0.5at^2

Answer 1

The correct equation to calculate the initial velocity during a vertical jump test, given the start and peak heights of the mass's center, is the kinetic energy equation vf^2 = vi^2 + 2a(rf - ri), with vf set to zero at the peak of the jump.

Step-by-step explanation:

To calculate the subject's initial velocity at takeoff during the vertical jump test, the correct equation to use is:

vf2 = vi2 + 2a(rf - ri)

Where vf is the final velocity, vi is the initial velocity, a is the acceleration (in this case, the acceleration due to gravity, which is approximately -9.81 m/s2 when directed downward), rf is the final position of the mass's center, and ri is the initial position of the mass's center. Since the subject's center of mass starts at 1.2 m and reaches 3.0 m at the peak, and knowing that the final velocity at the peak is 0 m/s (since the subject stops ascending at the peak), we can plug in the values to solve for the initial velocity.