Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m(4 ft). (This means that he moved upward by 1.2m after his feet left the floor.) Griffith weighed 890 N(200lb). (a) What was his speed as he left the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his average acceleration (magnitude and direction) while he pushed against the floor? (c) Draw his free-body diagram. In terms of the forces on the diagram, what was the net force on him? Use Newton’s laws and the results of part (b) to calculate the average force he applied to the ground.

The given data can be listed below as:

• Darrell Griffith moved upward to a distance of s=1.2 m.
• The weight of Griffith is $F_g=890N$.

The third equation of motion is used for identifying the motion of a particle. The third equation of motion is described as the square of the final velocity that equals the addition of the square of the initial velocity and two times of the product of acceleration and distance.

(a)

The equation of the speed of the basketball player can be expressed as:

$v^2=u^2+2gs$

Here, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and s is the distance moved by the player.

Substitute 0 for v,$-9.8m/s^2$ forg and 1.2 m fors in the above equation.

$$\begin{gathered} 0={\mathrm{u}}^2+2\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.2\mathrm{m}\right) \\ {\mathrm{u}}^2=\left(19.6\mathrm{m}/{\mathrm{s}}^2\right)\left(1.2\mathrm{m}\right) \\ =23.52{\mathrm{m}}^2/{\mathrm{s}}^2 \\ =4.84\mathrm{m}/\mathrm{s} \end{gathered}$$

The negative value is neglected as the velocity cannot be negative.

Thus, his speed as he left the floor is 4.84 /s.

(b)

The equation of the average acceleration is expressed as:

$a=\frac{u-v}{t}$

Here,a is the average acceleration andt is the time of the part of the jump.

Substitute 4.84 m/s for u, 0 for v and 0.300 s for t in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{4.84\mathrm{m}/\mathrm{s}-0}{0.300\mathrm{s}} \\ =\frac{4.84\mathrm{m}/\mathrm{s}}{0.300\mathrm{s}} \\ =16.17\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

As the acceleration is positive, hence it is directed upwards.

Thus, his average acceleration while he pushed against the floor is$16.17\mathrm{m}/{\mathrm{s}}^2$ in upward -direction.

The free body diagram of the player has been drawn below:

In the above diagram, the acceleration$a_{av,y}$ is directed upward. The player exerts a downward force and the normal force exerted on the player is $F_N$.

The equation of the net force on the player is expressed as:

$F_N=F_g+\frac{F_g}{g}{a}_{av,y}$

Here,$F_N$ is the net force,$F_g$ is the weight of Griffith and$a_{av,y}$ is the average acceleration.

Substitute 890 N for $F_g$,$9.8\mathrm{m}/{\mathrm{s}}^2$ for g and$16.17\mathrm{m}/{\mathrm{s}}^2$for ${\mathrm{a}}_{\mathrm{av},\mathrm{y}}$in the above equation.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{N}}=890\mathrm{N}+\frac{890\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}\left(16.17\mathrm{m}/{\mathrm{s}}^2\right) \\ =890\mathrm{N}+890\mathrm{N}\left(1.65\right) \\ =890\mathrm{N}+1468.5\mathrm{N} \\ =2358.5\mathrm{N} \end{gathered}$$

Thus, the net force on him is 2358.5 N.

According to the third law of Newton, the normal force applied by the player on the ground will be equal in magnitude but opposite in direction of the average force applied by the player. Hence, the average force of the player is -2358.5 N.

Thus, the average force he applied on the ground is -2358.5N .