A high-jump athlete leaves the ground, lifting his center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m/s. At what minimum speed must she leave the ground to accomplish this?

the minimum speed she must leave the ground with must be 6.1 m/s.

Explanation:

Since there are no external forces acting the the athlete, we can consider this as an isolated system. We will use the conservation of mechanical energy which states that the initial mechanical energy of a system is equal to the final mechanical energy of the system.

E_i = E_f

EK_i + EP_i = EK_f + EP_f

1/2 m v_i² + m g h_i = 1/2 m v_f² + m g h_f (1)

We are given:

v_f = 1.4 m/s h_f = 1.8 m

We want to determine:

v_i

Therefore, rearranging equation (1) to make v_i the subject of the equation, we get:

v_i = √[ v_f² + 2 (g) (h_f) ]

v_i = √[ (1.4 m/s) ² + 2 (9.8 m/s²) (1.8 m) ]

v_i = 6.1 m/s

Therefore, the minimum speed she must leave the ground with must be 6.1 m/s.