Jesse Owens physics...?

Question

A film of Jesse Owens's famous long in the 1936 Olympics shows that his center of mass rose 1.1 m from launch point to the top of the arc. What minimum speed did he need at launch if he was traveling at 6.5 m/s at the top of the arc. Please tell me HOW to get to the answer cuz I dont even know how to start it...

Answer 1

You must compute the total energy of the jumper both at the top of the jump and at the launch point. His energy at the top has two components: kinetic energy and potential energy. At the launch he has only kinetic energy. When you equate the two energy values, you can 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 .5*m*vt^2 + m*g*h
Energy at launch = .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 = √[vt^2 + 2*g*h]