Chapter 3, Problem 69GP

To determine

The range of initial speeds allowed to make the basket.

Answer to Problem 69GP

The minimum speed is $10.76\text{m/s}$ and the maximum speed is $10.96\text{m/s}$ .

Explanation of Solution

Given:

A basket ball leaves a player’s hand at a height of 2.10 m above the floor.

The distance between basket and floor is 2.60 m.

The player shoot the ball at a ${38.0}^{\text{o}}$ angle.

A shot is made from the horizontal distance of 11.0 m with an accuracy of $\pm 0.22\text{m}$ .

Formula used:

Apply the second equation of the motion to determine the ranges of initial speeds. The formula is

  $\mathrm{s}=\mathrm{u}\mathrm{t}+\frac{1}{2}\mathrm{a}{\mathrm{t}}^2$

Where, s is the distance, $\mathrm{u}$ is the velocity, $\mathrm{t}$ is the time and $\mathrm{a}$ is the acceleration.

Calculation:

At a constant speed, the projectile will move with an acceleration, $\mathrm{a}=0$ .

Le the player will travel a distance s_d in time t .

The distance is

  $\begin{array}{c}{\mathrm{s}}_{\mathrm{d}}=11\pm 0.22\\ =11+0.22\text{and}11-0.22\\ =11.22\text{m}\text{and}10.78\text{m}\end{array}$

Minimum distance is

  ${\mathrm{s}}_1=10.78\text{m}$

Maximum distance is,

  ${\mathrm{s}}_2=11.22\text{m}$

When the motion is in horizontal direction, calculate distance using the formula,

  ${\mathrm{s}}_{\mathrm{d}}=\mathrm{u}\mathrm{t}$

  $\begin{array}{l}{\mathrm{s}}_{\mathrm{d}}=\mathrm{u}\cos \left(38°\right)\mathrm{t}\\ \mathrm{t}=\frac{{\mathrm{s}}_{\mathrm{d}}}{\mathrm{u}\cos \left(38°\right)}\end{array}$

  

Figure 1

When the motion is in vertical direction, under the constant acceleration due to gravity, then the distance travelled by projectile is,

  $\begin{array}{l}\mathrm{s}=2.6-2.1\\ \mathrm{s}=0.5\text{m}\end{array}$

Now, use the formula,

  $\mathrm{s}=\mathrm{u}\mathrm{t}+\frac{1}{2}\mathrm{a}{\mathrm{t}}^2$

Substitute the values,

  $\begin{array}{l}0.5=\mathrm{u}\sin \left(38°\right)\frac{{\mathrm{s}}_{\mathrm{d}}}{\mathrm{u}\cos \left(38°\right)}+\frac{1}{2}\left(-\mathrm{g}\right){\left(\frac{{\mathrm{s}}_{\mathrm{d}}}{\mathrm{u}\cos \left(38°\right)}\right)}^2\\ 0.5={\mathrm{s}}_{\mathrm{d}}\tan \left(38°\right)-\frac{\mathrm{g}{\mathrm{s}}_{\mathrm{d}}{}^2}{2{\mathrm{u}}^2{\cos }^2\left(38°\right)}\end{array}$

Further solving the terms,

  $\begin{array}{c}\frac{\mathrm{g}{\mathrm{s}}_{\mathrm{d}}{}^2}{2{\mathrm{u}}^2{\cos }^2\left(38°\right)}={\mathrm{s}}_{\mathrm{d}}\tan \left(38°\right)-0.5\\ {\mathrm{u}}^2=\frac{\mathrm{g}{\mathrm{s}}_{\mathrm{d}}{}^2}{2{\cos }^2\left(38°\right)\left({\mathrm{s}}_{\mathrm{d}}\tan \left(38°\right)-0.5\right)}\end{array}$

Substitute 10.78 m for s_d and $9.8\text{m}/{\text{s}}^{\text{2}}$ for g to find the minimum velocity.

  $\begin{array}{l}{\mathrm{u}}^2=115.867918\\ \mathrm{u}=10.76\text{m/s}\end{array}$

Substitute 11.22 m for s_d and $9.8\text{m}/{\text{s}}^{\text{2}}$ for g to find the maximum velocity.

  $\begin{array}{l}{\mathrm{u}}^2=120.29947\\ \mathrm{u}=10.96\text{m/s}\end{array}$

Conclusion:

Therefore, the minimum range is $10.76\text{m/s}$ and the maximum range is $10.96\text{m/s}$ .