# Calculating Missile Launch Time for 2D Kinematics Problem

• arb
In summary, the problem involves an antitank gun on a 60m high plateau firing at an enemy tank located 2200m away. The tank starts from rest and accelerates at 0.5m/s^2 while the gun is fired at an angle of 10 degrees with a speed of 240m/s. To determine the time the gun should be fired in order to hit the tank, the projectile's position and the tank's position should be calculated using the equations V = V(initial) + at and r = r(initial) + v(initial)*t + 1/2at^2 for both the x and y components. However, the correct approach for solving this problem involves considering the relative velocity of the
arb
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)

sankalpmittal said:
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)

arb said:
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.

Arka420 said:
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.

## 1. What is 2D Kinematics?

2D Kinematics is the branch of physics that deals with the motion of objects in two-dimensional space, taking into account their position, velocity, and acceleration over time.

## 2. What are the key equations used in 2D Kinematics?

The key equations used in 2D Kinematics include the equations for displacement, velocity, and acceleration in both the x and y directions, as well as the Pythagorean theorem and trigonometric functions for finding the magnitude and direction of a vector.

## 3. How do you solve a 2D Kinematics problem?

To solve a 2D Kinematics problem, you first need to identify the known and unknown variables, and then choose an appropriate equation to solve for the unknown. It is important to carefully draw a diagram and label all given information. You may also need to use vector addition and trigonometry to find the magnitude and direction of a vector.

## 4. Can you use 2D Kinematics to solve real-world problems?

Yes, 2D Kinematics can be used to solve real-world problems such as calculating the trajectory of a projectile, predicting the motion of objects in a roller coaster, or analyzing the movement of a car in a two-dimensional plane.

## 5. What are some limitations of 2D Kinematics?

2D Kinematics assumes that all motion occurs in a flat, two-dimensional plane, which is not always the case in the real world. It also does not take into account air resistance, friction, or other external forces that may affect the motion of objects. Additionally, 2D Kinematics only applies to objects moving with constant acceleration, so it cannot be used for objects with changing accelerations.

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