- #1

FlukeATX

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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!