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

Answer:

7.037 m/s

Explanation:

According to the law of energy conservation;

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

This expression implies that:

$\dfrac{1}{2}mv^2 = mgh + \dfrac{1}{2}mu^2$

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

∴

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

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

$v = \sqrt{ 49.5124 \ m^2/s^2}$

v = 7.037 m/s