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Q90P

Expert-verifiedFound in: Page 90

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angle${{\mathbf{\theta}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{55}}{\mathbf{\xb0}}$above the horizontal, to make the foul shot? The horizontal distances are ${{\mathbf{d}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{ft}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{d}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{14}}{\mathbf{}}{\mathbf{ft}}{\mathbf{,}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{\mathbf{the}}{\mathbf{}}{\mathbf{heights}}{\mathbf{}}{\mathbf{are}}{\mathbf{}}{{\mathbf{h}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{7}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{ft}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{h}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{ft}}{\mathbf{.}}$**

The initial speed of the basketball is 23 ft/s

$\mathrm{a})\mathrm{The}\mathrm{angle}\mathrm{made}\mathrm{by}\mathrm{the}\mathrm{basketball}\mathrm{above}\mathrm{the}\mathrm{horizontal},\left(\mathrm{\theta}\right)=55\xb0\phantom{\rule{0ex}{0ex}}\mathrm{b})\mathrm{The}\mathrm{horizontal}\mathrm{distances},{\mathrm{d}}_{1}=1\mathrm{ft},{\mathrm{d}}_{2}=14\mathrm{ft}\phantom{\rule{0ex}{0ex}}\mathrm{c})\mathrm{The}\mathrm{heights}\mathrm{are},{\mathrm{h}}_{1}=7\mathrm{ft},{\mathrm{h}}_{2}=10\mathrm{ft}$

**Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration. (Upward is taken to be a positive direction.)**

**Using the equation for the projectile path in the projectile motion of the ball, we can find the initial speed of the basketball.**

Formula:

The equation of the distance for the projectile path of a body,

$\mathrm{y}=\mathrm{x}{\mathrm{tan\theta}}_{0}-\frac{{\mathrm{gx}}^{2}}{2{\left({\mathrm{V}}_{0}{\mathrm{cos\theta}}_{0}\right)}^{2}}$ …(i)

The total horizontal displacement of basketball is given as:

$\mathrm{x}={\mathrm{d}}_{2}-{\mathrm{d}}_{1}\phantom{\rule{0ex}{0ex}}=13\mathrm{ft}$

The total vertical displacement of basketball is given as:

$\mathrm{y}={\mathrm{h}}_{2}-{\mathrm{h}}_{1}\phantom{\rule{0ex}{0ex}}=3\mathrm{ft}$

Rearranging the equation (i)for projectile path and substituting the above values, the velocity is given as:

${\mathrm{V}}_{0}=\frac{\mathrm{x}}{\mathrm{cos\theta}}\sqrt{\frac{\mathrm{g}}{2\left(\mathrm{x}\mathrm{tan\theta}-\mathrm{y}\right)}}\phantom{\rule{0ex}{0ex}}=\frac{13\mathrm{ft}}{\mathrm{cos}55\xb0}\sqrt{\frac{32.15\mathrm{ft}/{\mathrm{s}}^{2}}{2\left(\left(13\mathrm{ft}\right)\mathrm{tan}55\xb0-\left(3\mathrm{ft}\right)\right)}}\phantom{\rule{0ex}{0ex}}=\frac{13\mathrm{ft}}{0.574}\sqrt{\frac{32\mathrm{ft}/{\mathrm{s}}^{2}}{31.13\mathrm{ft}}}\phantom{\rule{0ex}{0ex}}=22.98\mathrm{ft}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\approx 23\mathrm{ft}/\mathrm{s}$

Hence, the value of the initial speed of the ball is 23ft/s.

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