Finding distance and angle of crash between plane and rocket

[jheld]

Homework Statement

In a military test, a 575 unmanned spy plane is traveling north at an altitude of 2700 and a speed of 450 . It is intercepted by a 1280 rocket traveling east at 725 .

If the rocket and the spy plane become enmeshed in a tangled mess, where, relative to the point of impact, do they hit the ground? Give the direction as an angle east of north.

Homework Equations

F = ma
p = sqrt((m_1v_1)^2+(m_2v_2)^2)
vinitial = p/sum of m
sf = si + vi*t +1/2a*t^2

The Attempt at a Solution

I found Vinital by doing v = p/m, and got 519.352 m/s.

then i plugged that value into yf = yi + vi*t + 1/2a*t^2, and using yf = -2700.

then i used the quadratic formula to solve for t, and using the negative root (because of logistics) and ended up with t = 4.966 seconds.
then plugged that into xf = xi + vi*t, and ended up with 2.58 km.
the answer should be 12.2 km. so, I'm not sure what I am doing incorrectly.

 

[LowlyPion]

Homework Statement

In a military test, a 575 unmanned spy plane is traveling north at an altitude of 2700 and a speed of 450 . It is intercepted by a 1280 rocket traveling east at 725 .

If the rocket and the spy plane become enmeshed in a tangled mess, where, relative to the point of impact, do they hit the ground? Give the direction as an angle east of north.

Homework Equations

F = ma
p = sqrt((m_1v_1)^2+(m_2v_2)^2)
vinitial = p/sum of m
sf = si + vi*t +1/2a*t^2

The Attempt at a Solution

I found Vinital by doing v = p/m, and got 519.352 m/s.

then i plugged that value into yf = yi + vi*t + 1/2a*t^2, and using yf = -2700.

then i used the quadratic formula to solve for t, and using the negative root (because of logistics) and ended up with t = 4.966 seconds.
then plugged that into xf = xi + vi*t, and ended up with 2.58 km.
the answer should be 12.2 km. so, I'm not sure what I am doing incorrectly.

There is no quadratic. There is merely the time for it to fall from 2700 m

2700 = 1/2*(9.8)*t²

It's this time times your horizontal velocity that determines how far away.

 

[jheld]

Okay, I understand that in doing this, you get the 12.2 km, but isn't the equation:
yfinal = yinital + vinitial*t + 1/2*a*t^2?
so why cut off part of the equation?

thanks for the help

 

[LowlyPion]

Okay, I understand that in doing this, you get the 12.2 km, but isn't the equation:
yfinal = yinital + vinitial*t + 1/2*a*t^2?
so why cut off part of the equation?

thanks for the help

Strictly speaking the the plane and missile are in a horizontal x,y plane. Whatever velocity they have is not in the z-axis (up/down). Since there are no components of velocity from the crash in the z direction your initial z-velocity is 0, but it is the only component subject to gravity.

Hence the equation devolves into the simple z = 1/2*g*t²