91Ó°ÊÓ

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,

$\begin{array}{c}0−u^2=2×\left(−9.8\right){\text{ m/²õ}}^{\text{2}}×0.9\text{ m}\\ u=\sqrt{17.64{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}}\\ u=4.2\text{ m/²õ}\end{array}$

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,

$\begin{array}{c}{4.2}^2{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}−0=2×a×0.3\text{ m}\\ 17.64{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}−0=a×0.6\text{ m}\\ a=\frac{17.64{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}}{0.6\text{ m}}\\ a=29.4{\text{ m/²õ}}^2\end{array}$

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,

$\begin{array}{c}F−110\text{ k²µ}×{\text{9.8 m/²õ}}^2=110\text{ k²µ}×{\text{29.4 m/²õ}}^2\\ F−1078\text{ k²µ}â‹\dots {\text{m/s}}^2=3234\text{ k²µ}â‹\dots {\text{m/s}}^2\\ F=\left(3234+1078\right)\text{ N}\\ \text{F}=4312\text{ N}\end{array}$

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