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Fundamentals Of Physics
Found in: Page 85

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Short Answer

A certain airplane has a speed of290.0 km/hand is diving at an angle ofθ=30.0°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 isd=700 m(a) how long is the decoy in the air? (b)How high was the release point?

(a) The decoy is in the air for10.0 s

(b) The release point was897 mhigh.

See the step by step solution

Step by Step Solution

Step 1: Given information

Initial speed of airplane is

v0=290kmhr =80.6ms

Projection angle isθ=-30° below the horizontal.

Horizontal distance between release point of the decoy and the point where decoy strike the ground isdx=700 m

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.


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

dx=(v0cosθ)t (i)



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

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

t=dxv0cosθt=700 m80.6 m/scos-30.0°t=10.02842 s 10.0 s

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


dy-dy0=v0sinθt-12gt2As dy=0 m,dy-dy0=80.6sin-30.0°10.02842 s-129.8ms210.02842s2dy0=896.934 m 897 m

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