Kinematics in 2d, plane drops an object

  • Thread starter FlukeATX
  • Start date
  • #1
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Homework Statement


A certain airplane has a speed of 267.2 km/h and is diving at an angle of θ = 28.0° below the horizontal when the pilot releases a radar decoy (Fig. 4-37). The horizontal distance between the release point and the point where the decoy strikes the ground is d = 777 m. (a) How long is the decoy in the air? (b) How high was the release point?


Homework Equations


d=Vi*t+1/2a(t^2)


The Attempt at a Solution


Okay, step one is to find the velocity in the x and y directions, here is my attempt:
Vy=sin(-28)*267.2= -125.4428016
Vx=cos(-28)*267.2= 235.9235968

So we have the following knowns:
V=267.2m/s
Vy=-125.4428016m/s
Vx=235.9235968m/s
Dx=777m
Ax=0
Ay=-9.8m/s^2
Dy=???
T=???

Okay, cool, plenty of information. Let's solve for T first so we can find Dy afterwards:
Dx=VxT + 1/2Ax(T^2)
777=(235.9235968)(T) + (1/2)0(T^2)
777=(235.9235968)(T) + 0
T=777/(235.9235968)
T=3.293439107

----STOP----
Now at this point, I went on to solve for Dy, but we're gonna stop here for the following reason: when I submitted my answer for T to the wileyplus online system, it told me my answer was wrong. What did I do wrong? Everything seems to make sense. Please help!
 

Answers and Replies

  • #2
Nabeshin
Science Advisor
2,205
16
The speed of the airplane is given in km/hr. You should check for consistency in your units ;)
 
  • #3
7
0
Ah jeez, totally overlooked that. Thanks for the help!
 

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