Question

In a film of jesse owens's famous long jump in the 1936 olympics, it is observed that his center of mass rose 1.02 m from launch point to the top of the arc. what minimum speed did he need at launch if he was also noted to be travelling at 6.35 m/s at the top of the arc? round your answer to the second decimal place.

Answer 1

To find the minimum speed needed at launch in Jesse Owens' long jump, we can use the conservation of mechanical energy. Using the given information, the initial speed is calculated to be 10.77 m/s.

Step-by-step explanation:

To determine the minimum speed needed by Jesse Owens at launch, we can use the concept of conservation of mechanical energy. The change in potential energy (due to the rise of the center of mass) is equal to the change in kinetic energy.

Since the center of mass rose 1.02 m and the speed at the top of the arc is 6.35 m/s, we can calculate the initial speed using the equation:

initial speed = final speed + sqrt(2 * acceleration due to gravity * height)

By substituting the given values, we can calculate the initial speed:

initial speed = 6.35 m/s + sqrt(2 * 9.8 m/s^2 * 1.02 m) = 6.35 m/s + 4.42 m/s = 10.77 m/s