Question

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.50 m and cross the bar with a speed of 0.68 m/s

Answer 1

7.037 m/s

Step-by-step explanation:

According to the law of energy conservation;

The total of the initial K.E = The total final K.E

This expression implies that:

`(1)/(2)mv^2 = mgh + (1)/(2)mu^2`

By simplifying the above expression;

`v^2 = 2gh + u^2`

where;

g = 9.81 m/s²

h = 2.50 m

u = 0.68 m/s

∴

`v^2 = 2(9.81 \ m/s^2 )(2.50 \ m) + (0.68 \ m/s)^2`

`v^2 = 49.5124 \ m^2/s^2`

`v = √( 49.5124 \ m^2/s^2)`

v = 7.037 m/s