Question

Two warlords aim identical catapults (i.e. they both release rocks at the same speed) at each other, with both of them being at the same altitude. The warlords have made the necessary computations to crush the other and fire their catapults simultaneously. Amazingly, the two stones do not collide with each other in mid-air, but instead, the stone Alexander fired passes well below the stone that Genghis shot. Genghis is annihilated 8.0s after the catapults are fired, and Alexander only got to celebrate his victory for 4.0s before he too was destroyed. Find the amount of time that elapses from the launch to the moment that the rocks pass each other in the air.

Answer 1

Alright, let's see how this plays out. Genghis and Alexander shoot at the same time with the same initial velocity . They both have the same range, but different travel times, such that Alexander's shot reaches Genghis first. From intuition it is apparent that Genghis shot higher and Alexander lower. So the difference in the initial angle for Genghis and Alexander respectively is what decides everything. How does that work? Let's take a normal projectile motion example.

This is the general equation of path for a projectile thrown with velocity and angle . Range of this projecticle, which is the horizontal distance covered before reaching the same initial height,

The time taken for it to travel this distance, will be, which be the total time of flight.

Now, the reason for projectile motion is simple. Unlike horizontal motion where there is only one component of velocity , a projectile is thrown with some angle to the horizontal x-axis. Here, the x-component of velocity will remain constant, and the y-component of velocity will change as the particle rises vertically and faces slow down due to gravitional decceleration. We will ignore air resistance for simplicity.

So, just like a ball thrown straight up, here too the y-component of velcoity will decrease till it becomes 0 at some time t and height H. Now the object will accelerate downwards, negatively decreasing the y-component of velocity till again it reaches the ground.

Now, a ball thrown up with velocity v will take the same time to reach its maximum height as the time it takes to then fall down again. From that we can stipulate that here too, when the time is , the particle will have reached its maximum height (verticle distance from the ground) .

One last thing before we solve our problem, take a look at equation of range R. It has a sine fucntion, which we know varies within the limits of . Since, the maximum value of assuming +ve axis in the dierection of the throw, the values of will remain in the 1st and 2nd quadrant going from 0 to 1 to 0 again. From this we can assume there has to be a maximum value for the range of the object thrown after which the increase in angle starts to reduce the range again. That value occurs at,

So there is symmetry in the motion, for above and below the angle of 45 degrees, such that,

Using this fact, the range of Genghis and the Alexander being the same, their projection angles were just some value above and below 45.

We'll use this later.

Now, we know that it took 8 seconds for Genghis to be hit and 8+4 = 12 seconds for Alexander to be hit.

So their total time of flight are,

Since Alexander's shot was faster and at a lower angle, both shots would have met to the left of half of their range. At this point suppose the shot from Genghis would have traveled and Alexdander's shot would have traveled .

The horizontal position is given by,

Dividing both equations,

We can also find the range and the initial velocity for the shots and since they are same we can use just Genghis' value which gives us,

Substituting in equation (A),

And from this can get the time taken for the shot to reach this point,