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Q4.7-46PE

Expert-verifiedFound in: Page 164

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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.**

(a) The initial velocity of the player is 4.2 m/s.

(b) The acceleration is 29.4 m/s^{2}.

(c) The force exerted on the floor is -4312 N

- Hight = 0.900 m.

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/s^{2} 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/s ^{2}.**

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/s^{2} for *a*, and 9.8 m/s^{2} 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.**

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