Kinematics in 2d, plane drops an object

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SUMMARY

The discussion centers on a physics problem involving kinematics, where an airplane traveling at 267.2 km/h releases a radar decoy at a 28.0° angle. The horizontal distance to the impact point is 777 m. The user calculated the horizontal (Vx) and vertical (Vy) components of the velocity, arriving at Vx = 235.92 m/s and Vy = -125.44 m/s. A critical error identified was the inconsistency in units, as the airplane's speed was initially given in kilometers per hour rather than meters per second, which led to an incorrect calculation of the time the decoy was in the air.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = Vi*t + 1/2a(t^2)
  • Knowledge of vector decomposition to find horizontal and vertical components of velocity
  • Familiarity with unit conversions, particularly between kilometers per hour and meters per second
  • Basic grasp of gravitational acceleration (Ay = -9.8 m/s²)
NEXT STEPS
  • Learn how to convert speed from kilometers per hour to meters per second
  • Study the application of kinematic equations in projectile motion
  • Explore vector decomposition techniques for analyzing motion in two dimensions
  • Investigate common pitfalls in physics problems related to unit consistency
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for examples of common mistakes in problem-solving.

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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 going to 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!
 
Physics news on Phys.org
The speed of the airplane is given in km/hr. You should check for consistency in your units ;)
 
Ah jeez, totally overlooked that. Thanks for the help!
 

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