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Q27P

Expert-verifiedFound in: Page 85

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A certain airplane has a speed of${}^{\mathbf{290}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{k}\mathbf{m}\mathbf{/}\mathbf{h}}$and is diving at an angle of${}^{\mathbf{\theta}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{\xb0}}$below the horizontal when the pilot releases a radar decoy (Fig. 4-33). The horizontal distance between the release point and the point where the decoy strikes the ground is${}^{\mathbf{d}\mathbf{=}\mathbf{700}\mathbf{}\mathbf{m}}$(a) how long is the decoy in the air? (b)How high was the release point?**

(a) The decoy is in the air for${}^{10.0s}$

(b) The release point was${}^{897m}$high.

Initial speed of airplane is

${\mathrm{v}}_{0}=290\frac{\mathrm{km}}{\mathrm{hr}}\phantom{\rule{0ex}{0ex}}=80.6\frac{\mathrm{m}}{\mathrm{s}}$

Projection angle is${}^{\theta =-30\xb0}$ below the horizontal.

Horizontal distance between release point of the decoy and the point where decoy strike the ground is${}^{{d}_{x}=700m}$

**This problem deals with kinematic equations that describe the motion of an object with constant acceleration. Using the standard equation for the velocity of the object, the time that the decoy spent in air can be computed. Further, using the formula for the second kinematic equation, the height of release point can be found.**

Formula:

The displacement for the horizontal and vertical direction can be written as,

${d}_{x}=\left({v}_{0}\mathrm{cos}\theta \right)t$ (i)

${d}_{y}-{d}_{y0}=\left({v}_{0}\mathrm{sin}\theta \right)t-\frac{1}{2}g{t}^{2}$

(ii)

Now, using equation (i) the time will be,

$\mathrm{t}=\frac{{\mathrm{d}}_{\mathrm{x}}}{{\mathrm{v}}_{0}\mathrm{cos\theta}}\phantom{\rule{0ex}{0ex}}\mathrm{t}=\frac{700\mathrm{m}}{\left(80.6\mathrm{m}/\mathrm{s}\right)\mathrm{cos}\left(-30.0\xb0\right)}\phantom{\rule{0ex}{0ex}}\mathrm{t}=10.02842\mathrm{s}\phantom{\rule{0ex}{0ex}}\approx 10.0\mathrm{s}$

Now,

${\mathrm{d}}_{\mathrm{y}}-{\mathrm{d}}_{\mathrm{y}0}=\left({\mathrm{v}}_{0}\mathrm{sin\theta}\right)\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{As}{\mathrm{d}}_{\mathrm{y}}=0\mathrm{m},\phantom{\rule{0ex}{0ex}}{\mathrm{d}}_{\mathrm{y}}-{\mathrm{d}}_{\mathrm{y}0}=\left(80.6\right)\mathrm{sin}\left(-30.0\xb0\right)\left(10.02842\mathrm{s}\right)-\frac{1}{2}\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^{2}}\right){\left(10.02842\mathrm{s}\right)}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{d}}_{\mathrm{y}0}=896.934\mathrm{m}\phantom{\rule{0ex}{0ex}}\approx 897\mathrm{m}$

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