In a film of Jesse Owens\'s famous long jump in the 1936 Olympics, it is observe
ID: 3893823 • Letter: I
Question
In a film of Jesse Owens's famous long jump in the 1936 Olympics, it is observed that his center of mass rose 1.07m from launch point to the top of the arc. What minimum speed did he need at launch if he was also noted to be traveling at 6.53m/s at the top of the arc? I need a final answer at the end please Thanks In a film of Jesse Owens's famous long jump in the 1936 Olympics, it is observed that his center of mass rose 1.07m from launch point to the top of the arc. What minimum speed did he need at launch if he was also noted to be traveling at 6.53m/s at the top of the arc? I need a final answer at the end please ThanksExplanation / Answer
His energy at the top has two components: kinetic energy and potential energy. At the launch he has only kinetic energy. equating the two energy values, to find the initial velocity at launch.
Kinetic energy = .5*m*v^2
Potential energy = m*g*h (h is the elevation at the top of the jump)
Therefore
At the top the total energy is 0.5*m*vt^2 + m*g*h
Energy at launch = 0.5*m*vl^2
vt = velocity at top, vl = velocity at launch
Equating these gives
.5*m*vl^2 = .5*m*vt^2 + m*g*h
The mass m appears in every term so it can be divided out, leaving
.5*vl^2 = .5*vt^2 + g*h
Solve for vl:
vl = sqrt([vt^2 + 2*g*h]) = 7.976 m/sec
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