Physics Question...?

Question

Yes, this is a homework question! :) A projectile that is fired from a gun has an initial velocity of 90km/h at an angle of 60degrees above the horizontal. When the projectile is at the top of it's trajectory, an internal explosion causes it to separate into two fragments of equal mass. One of the fragments falls straight downward as tough it had been released from rest. How far from the gun does the other fragment land? PLEASE HELP ME WITH THIS!! :)

Answer 1

You first have to find the components ( vertical, and horizontal or x and y ) by using sine and cosine of the angle. Once you have found these, everything sholud be pretty easy from there. Remember, the acceleration is due to gravity because the fragment that is traveling downward and the one still being projected will reach the horizon or the ground at the same time.

Answer 2

I agree with dutch prof. I didn't crank out the numbers, but I think the responder with the TA experience and degree neglected to take into consideration the conservation of momentum. He didn't offer a good reason for ignoring it either.

When I took physics, we learned that the conservation of momentum was a fundamental law. What I would do, is solve using regular kinematics like the dutchprof did, for the top of the trajectory. Then, freeze frame.

there's a projectile that has a horizontal velocity of 90km/h * cos 60 = 12.5 m/s (notice the unit change). It has momentum of 12.5 m/s * M (the mass of the projectile). Then it splits into two pieces each of equal mass. So the momentum of the two pieces should still be 12.5 * M. Since the projectile that falls straight down with an initial velocity of zero (the problem statement says "as though it had been released from rest"), all of the momentum should be transferred to the other half of the projectile. Since it has half the mass, its velocity must double to 25 m/s.

Now you have half the projectile moving twice as fast in the horizontal direction. As the two previous responders found, it took 2.21 seconds to reach its pinnacle, so it will take the same amount of time to reach the ground seeing as it's the same vertical distance.

For the first 2.21 seconds of its journey, it traveled 2.21 s * 12.5 m/s = 27.625 meters.

On the way down, it traveled 2.21 s * 25 m/s = 55.25 meters.

It traveled a total of 82.875 meters.

Incidentally, when the projectile reaches the pinnacle, it only has horizontal momentum. There is no vertical momentum. That's why the piece that falls to the ground necessarily has an initial velocity of zero.

I may be wrong on this, but I'd use this or dutchprof's answer until someone explains why you shouldn't take momentum into consideration at all.

Answer 3

The first thing I would suggest is to draw a picture and find the verticle and horizontal components of the initial velocity. I have done the problem in units of meters/sec instead of km/hr. We find that 90 km/hr is exactly 25 m/s (90km/hr *1000m/km * 1hr/3600s=25 m/s). The verticle component is what we are most interested in now, we want to know how long it was in the air. We take 25 m/s and multiply by the sine of the angle 60 degrees to find the verticle component (25 m/s * SIN[60 degrees]~21.6 m/s).

They get a little tricky with this splitting into two parts business. This is information that is irrelevant to the answer unless the explosion changed the trajectory of the the fragment that does not fall straight down, which is what we are interested in. Since the problem says nothing about this, I will assume that the trajectory remains unchanged. Since I have decided to use a purely kinematics approach (no energy or force equations) mass is irrelevant to the final answer.

We will use the equation x=v0*t+1/2*a*t^2 to find the time of flight. I have set x=0m for the hieght of the ground v0=21.6m/s and a=-9.8m/s^2 and solved for t. I get two solutions for t (t=0 and t=4.4sec.) t=0 means that at the beginning time (time of launch) the projectile is at ground level, which makes sense and that its time of flight was when it hit the ground next at 4.4 sec.

We then find the horizontal component of velocity by taking the intial velocity of 25 m/s and multiplying by the cosine of 60 degrees to get 12.5 m/s. By taking the horizontal velocity, which remains constant, and multiplying the time of flight, 4.4 sec, we find how far it flew which is about 55.2m.

Answer 4

Now the trajectories. The initial velocity is 25 m/s, and decomposition gives

vx0 = 25 cos 60 = 12.5 m/s
vy0 = 25 sin 60 = 21.65 m/s

The velocity decreases according to the formula

vy = vy0 - gt = 21.65 - 9.81 t

so that the highest point is reached when

vy = 21.65 - 9.81 t = 0 --> t = 2.21 s

During that time, the projectile has traveled over a distance

x1 = vx . t = 12.5 x 2.21 = 27.59 m

Then the explosion takes place. The total momentum of the projectile is transferred to one fragment. Because the same momentum is now carried by half the mass, the (horizontal) velocity is double. It is now vx = 2 x 12.5 = 25 m/s.

The vertical motion for both fragments is the same as if the explosion had not happened. The falling time is equal to the time calculated before, t = 2.21 s.

The horizontal distance traveled by the fragment after the explosion is

x2 = vx . t = 25 x 2.21 = 55.17 m

The total distance is therefore

x1 + x2 = 82.76 m

Answer 5

Great question! I need to review. I know its not that hard. But first try thinking like this. Since one part fell straight down its Sin 100% and Cos is 0. Also since it says "like it was just dropped you don't have to worry about it. Thus all the energy is in the Cos direction is in the second half. Mass is not a problem thus treat it as if the round did not explode. I think that will work.

Answer 6

This too much for me, but the mass having a certain amount of force applied to it produces the temporary velocity of 90KmH. Power measured in grams*centimeter*sec for a unknown duration produces the velocity 90KmH for object mass M. Gravity power for this object is 16 feet (convert to metric)*M*sec*sec. Assume the duration of this quantity of force applied to the object is one second/X', its velocity shall be greater or less than 90KmH. Find the duration that gives 90KmH and that gives you a horizontal distance, but not the 60 degree climb. Assuming 90 degree vertical climb equals zero horizontal distance, then zero plus Zero degree climb distance (horizontal)/90/Centimeter/sec for every one degree from the vertical, zero plus horizontal distance/90/centimeter/sec/x/sec times 30

(x equals the powering duration of the weapon discharge)

Deceleration would be 16 feet in centimeters/M/sec/sec at the end of X'sec
Total mass/90Km/3600secs
The distance for the other half of its mass is equal to the total distance for its total mass.

Sorry, I am experincing RAM over run.

Answer 7

Probably no more than a few feet if you got your whole head in front of the barrel.

Answer 8

It doesn't matter - you need to fire another round.

Answer 9

yeah...fire some more rounds hommie......