What is the height of the plane when the decoy was released?

[B-80]

[SOLVED] projectile shot downward

A certain airplane has a speed of 255.0 km/h and is diving at an angle of 30.0° below the horizontal when a radar decoy is released (see Figure 4-36). The horizontal distance between the release point and the point where the decoy strikes the ground is 700 m. Neglect air resistance.

so Velocity in meters per second is 255*.2777=70.83m/s
Velocity in X direction is 70.83sin60=61.34m/s

Figure 4-36

(a) How high was the plane when the decoy was released?

t=m/v = 700/61.34 = 11.41s
Y=Yo+volt+.5(a)(t^2)
Y=(0)+(0)(t)-4.9(11.41^2)
Y=638.1m which is wrong, can anyone help me out

 

[Doc Al]

t=m/v = 700/61.34 = 11.41s
Y=Yo+volt+.5(a)(t^2)
Y=(0)+(0)(t)-4.9(11.41^2)

The y-component of the the initial velocity is not zero.

 

[bel]

The initial vertical component of the velocity of the released item is not zero, due to the vis inertiae, since the aeroplane was traveling at two hundred and fifty-five kilometres per hour times the sine of thirty degrees vertically.

 

[B-80]

good point :p