# 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.

I apologize the correct answer from the answer key is 10.9s. I was looking at the incorrect problem, So then now I am completely lost.

Your expression for ##r_{tank}(t)## does not look correct. The problem says that the tank starts from rest and has an acceleration of ##0.5 ms^{-2}##.

I will use concept of relative velocity and acceleration in it and would consequently convert this 2-body problem to a one body. If what i say is obscure to you, then answer this:

What is relative velocity ?

(Please don't say final velocity minus initial velocity. :P)

I will use concept of relative velocity and acceleration in it and would consequently convert this 2-body problem to a one body. If what i say is obscure to you, then answer this:

What is relative velocity ?

(Please don't say final velocity minus initial velocity. :P)

You mean calculate the velocity of the missile with respect to the tank? Is that what you are suggesting? (By velocity I mean the horizontal component or the x-component)

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.

You have made a mistake. The projectile meets the tank at some point in between the initial point of the tank and the anti-tank gun. How can you assume that the tank has traveled 2200m? It should be either 2200-x or x(x is the distance where the projectile strikes the tank. Form two equations,and solve them to find the value of x.

You mean calculate the velocity of the missile with respect to the tank? Is that what you are suggesting? (By velocity I mean the horizontal component or the x-component)

Yup. By bringing tank VIRTUALLY to rest.