# 2D Kinematics Question

An antitank gun which fires missiles with a speed of 240m/s is located on a 60m high plateau overlooking the surrounding plain. The gun crew sights an enemy tank 2200m from the base of the plateau. Simultaneously, the tank crew sees the gun and starts to head directly away from the plateau. The tank starts from rest and accelerates at .5m/s^2. The gun is fired at an angle of 10 degrees. For the missile to hit the tank, how long after the tank starts to move should the gun be fired?

I assume I will be using V = V(initial)+at and r=r(initial) + v(initial)*t +1/2at^2 for both the x and y component.

Here is my attempt.

for the antitank gun:

rx = 0 + 240cos(10)t
ry = 0 = 60 +240sin(10)t – 4.9t^2
t = -1.21s, 9.76s, ignoring the negative for this problem.

to find final position of missile rx = 240cos(10)(9.76s) = 2306.81m

for the tank:
r(Tank) = 2200+0+.5t
2306.81 = 2200+0+.5t
t = 213.63 sec for the tank to get from where it started to where the missile will hit it.

Since it takes 9.76 for the missile to arrive I would have though it would need to take off 213.63 s – 9.76 s after the tank starts moving.

The answer sheet says 9.8sec , I’m assuming rounding up from the 9.76 missile flight time. Could someone explain the logic to me? Why should I stop at just the tank flight time.

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