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Found in: Page 85

Fundamentals Of Physics

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${}^{\mathbf{\theta }\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}}$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.

See the step by step solution

Step 1: Given information

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${}^{\theta =-30°}$ below the horizontal.

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

Step 2: To understand the concept of kinematic equations

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)

Step 3: (a) To find the time that the decoy spent in air

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

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

Step 4: (b) To find the release point of the decoy

Now,

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