Question

A 75-kg volleyball player performs a maximum-effort vertical jump with an initial velocity of 3 m/s. The player leaves the ground with an upright body position and also touches down with the same body position. Answer the following: ua A. How much will the player's center of mass (COM) travel in the air along the vertical direction during the ascending phase of the jump? Compute the maximum kinetic energy of the player during the jump. C. Compute the maximum potential energy of the player during the jump. Compute the minimum mechanical energy of the player during the jump. E. Find the vertical velocity of the player's COM at the apex of the jump (top). F. Find the horizontal velocity of the player's COM half way up. G. Find the vertical acceleration of the player's COM in the air. H. Compute the vertical linear momentum of the player right before she touches down. I. If the player wants to reach a complete stop in 0.5 seconds after touch-down during the landing motion, how much average force does she need to exert to the ground in the vertical direction? J. Compute the kinetic energy of the player at the end of the landing action.

Answer 1

Given,

mass of the player m=75Kg

initial velocity u= 3 m/s.

(a) The height is given by,

  $h=\frac{ u^2}{2g}=\frac{3^2}{2\times 9.8}=0.46m$

(b) The maximum kinetic energy of the player is at bottom position.

  $K=\frac{1}{2}mu^2=\frac{1}{2}\times 75\times 3^2=337.5J$

(c) The maximum potential energy is equal to kinetic energy at the start of jump.

$U=337.5J$

(D) Mechanical eergy is the sum of kinetic and potential energy. The minimum mechanical energy of the player during jump is equal to

$T=337.5J$

(E) The vertical component of velocity at top is zero.

(F) The horizontal velocity is zero at any point. He has only vertical velocity.

(G) The vertical accerleration of the palyer is -g , ie -9.8m/s^2.

(H) The vertical linear momentum before touching the ground is mu.

  $P=mu=75\times 3=225Kgm/s$