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An antiaircraft gun fires shells at 200 m/s at a 60° angle. An enemy plane flies directly toward the gun at 300 m/s, 495 m off the ground. How far away (horizontally) must the plane be when the gun fires for the shell to hit the plane?

First off, I know I get two solutions here. I believe it's because of the trajectory of the shell as it hits the plane on the way up or on the way down.

Knowns

v_ix = 200cos(60) = 100

v_iy = 200sin(60)

a_x = 0

a_y = -9.81

y=485 m

Unknowns

x

v_fy

v_fx

t

Equations

y_f = y_i + v_iy * t + .5at^2

x_f = x_i + v_ix * t + .5at2

I'm not sure if I really have to take into account the -300 m/s of the plane, but here's what I've tried so far:

I used the above equation for the y direction to get

485 = 200sin(60)t -4.9t^2

0 = -4.9t^2 + 200sin(60)t-485

I used the quadratic equation to get two times, 3.066 s and 32.2819 s. I tried fitting these into the x direction equation above, but no luck. I may have done something wrong up to this point, but I think that where I go next is where my problems are occurring. I ended up getting 306.61 for the first distance and 3228 for the second, obviously nothing there. Please help me with the rest of the problem!

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# Homework Help: 2-D Kinematics Problem

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