A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor.

(a) Calculate his velocity when he leaves the floor.

(b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m.

(c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

Apply the equation of motion as:

${\mathit{v}}^{\mathbf{2}}\mathbf{-}{\mathit{u}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{g}\mathit{h}$

Here, v is the final speed, u is the initial speed, g is the acceleration due to gravity, and h is the height attained.

Substitute 0 for v, 0.9 m for h, and (-9.8) m/s² for g in the above expression, and we get,

$\begin{array}{c}0-u^2=2\times \left(-9.8\right){\text{\hspace{0.33em}m/s}}^{\text{2}}\times 0.9\text{\hspace{0.33em}m}\\ u=\sqrt{17.64{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}}\\ u=4.2\text{\hspace{0.33em}m/s}\end{array}$

Hence, the initial velocity of the player is 4.2 m/s.

Apply the equation of motion as:

${\mathit{v}}^{\mathbf{2}}\mathbf{-}{\mathit{u}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{a}\mathit{s}$

Here, a is the acceleration, and s is the distance covered.

Substitute 0 for u, 4.2 m/s for v, and 0.3 m for s in the above expression, and we get,

$\begin{array}{c}{4.2}^2{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}-0=2\times a\times 0.3\text{\hspace{0.33em}m}\\ 17.64{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}-0=a\times 0.6\text{\hspace{0.33em}m}\\ a=\frac{17.64{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}}{0.6\text{\hspace{0.33em}m}}\\ a=29.4{\text{\hspace{0.33em}m/s}}^2\end{array}$

Hence, the acceleration is 29.4 m/s².

The free-body diagram is shown below:

Apply Newton’s Second Law of motion as:

$\begin{array}{c}{F}_{net}=ma\\ F-mg=ma\end{array}$

Here, m is the mass of the player, F is the force exerted, and F_net is the net force exerted.

Substitute 110 kg for m, 29.4 m/s² for a, and 9.8 m/s² for g in the above expression, and we get,

$\begin{array}{c}F-110\text{\hspace{0.33em}kg}\times {\text{9.8\hspace{0.33em}m/s}}^2=110\text{\hspace{0.33em}kg}\times {\text{29.4\hspace{0.33em}m/s}}^2\\ F-1078\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2=3234\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2\\ F=\left(3234+1078\right)\text{\hspace{0.33em}N}\\ \text{F}=4312\text{\hspace{0.33em}N}\end{array}$

Hence, the force exerted on the floor is-4312 N as the force will be in the opposite direction.