Projectile Question on Range

Question

A projectile is thrown with speed 50√2 m/s at angle of 45° with horizontal.

At the highest point it explodes into three fragments of equal mass of which one of the fragment retraces its path,the second one falls freely downwards.Then the range of the third part?

Answer 1

at the highest point the velocity will be 50√2 x sin 45=50m/s

so

m₁v₁+m₂v₂+m₃v₃=mv

one retrace so v₁=50

one free fall so v₂=0

v₃=?

(m/3). 50 +(m/3).v₃=m50

so v₃₌₁₀₀

time to the highest point:  v=gt so 50=10.t so t=5s

 

first 5s it went with 50m/s that will make 50x5=250m

second 5s it went with 100m/s that will make 100x5=500m

so total range is 250+500=750m

 

 

Answer 2

Hi Aman!

 

I have had the chance to look this problem over a little more closely, and I think the previous answer is almost completely correct, but makes one small, though important, error that significantly affects the result.

 

The projectile is launched with a speed of 50√2 at an angle of 45^o with the horizontal.  This means that its initial vertical velocity if (50√2)sin(45^o) = 50 m/s, and its horizontal velocity is (50√2)cos(45^o) = 50 m/s.  That horizontal velocity will be constant, since, by definition, the only acceleration on a projectile is gravitational acceleration (g), which is vertical.  So there is no horizontal acceleration, and thus no horizontal change in velocity (if air resistance is ignored, which we must assume here, since no information is given to put it in the solution).

 

At the peak of its flight, like any projectile, it has only its horizontal velocity (the vertical velocity has gone to zero at the peak).  At this point, it explodes, as described.  The explosion is an internal event, so momentum is conserved.

 

At its peak, because it has only horizontal velocity, there is no initial vertical momentum for the projectile.  Therefore, must have zero net momentum vertically after the explosion.  Because the part of the projectile that falls is said to "fall freely," I take that to mean it falls from rest, as if it were dropped.  So it has no vertical momentum right after the explosion, since its initial vertical velocity (right after the explosion) is zero.  The first piece goes back the way the projectile came, which was a horizontal path, so it has no vertical momentum right after the explosion, either.  Since there was no vertical momentum before the explosion, and the other two pieces have no vertical momentum right after, the third piece must have no vertical momentum right after the explosion, either.  Therefore, the third piece must have only horizontal momentum, and thus only horizontal velocity.

 

Now, let's look at momentum conservation through the explosion in the horizontal direction.  Initially, the projectile is moving horizontally (let's say in the positive direction) with that horizontal speed of 50 m/s.  So, its initial momentum is:

 

p_o = 50(3m) kg•m/s = 150m kg•m/s

 

where m is the (equal) mass of the fragments after the explosion.

 

The second piece falls freely, directly down, after the explosion.  So, in addition to have no vertical momentum right after the explosion, it also has no horizontal momentum then.  So its horizontal momentum after the explosion is 0.

 

The first piece retraces the projectile's path, which I take to mean that it goes back along the path from which the projectile came, which means it must be going at 50 m/s horizontally, as the projectile was.  Here is where I think the other solution is in error.  Momentum is a vector, so direction matters.  If we define the projectile to be going in the positive direction to start, and the one fragment goes back along the path the projectile came, it must be going at 50 m/s, but in the negative direction.  Hence, its horizontal momentum just after the explosion is -50m kg•m/s.

 

So the horizontal conservation of momentum equation becomes:

 

150m = -50m + 0 + m(v₃)

 

We can cancel the m's out, and this leaves:

 

v₃ = 200 m/s (in the positive direction)

 

Now we have to figure out the range (horizontal distance traveled).  Before the explosion, the projectile moves horizontally at 50 m/s for all the time between launch and reaching its highest point.  If we can calculate how long a time that is, we can calculate how far horizontally the projectile traveled before exploding.

 

The time to reach maximum height is determined by vertical direction motion.  So, using kinematics, we want to solve for time to max height in the vertical.

 

to find: t

know: a (= -9.8 m/s², taking down as negative), v_oy (= 50 m/s, calculated above), v (=0, since it is at maximum height)

 

So, we can use the kinematic equation that deals with these four quantities:

 

v = v_o + at   (this is actually just an algebraic rearrangement of the definition of acceleration)

 

Solve this for t.

 

0 = 50 m/s + (-9.8 m/s²)t

 

t = 5.1 s

 

Because the horizontal motion is at constant velocity (as indicated above), we can determine how far it travels horizontally by just using:

 

d = vt

d = (50 m/s)(5.1 s) = 255 m

 

Now, the part of the projectile that shoots forward at 200 m/s has no initial vertical velocity, so its vertical motion characteristics are exactly the same as if it had been dropped.  In projectiles, there is a symmetry to up and down motion.  If a launched object takes 5.1 s to reach a certain height, it will take 5.1 to get to the ground if dropped from that height.  We could demonstrate this, but I will just appeal to this symmetry argument, since this question already has enough in it :).  We can prove that, if you need to.

 

What this comes down to is that the piece that shoots forward with only horizontal velocity will also take 5.1 s to get back to ground.  In that time, it will travel:

 

d = v₃t = (200 m/s)(5.1 s) = 1020 m

 

Hence, the total range on the third piece (including when it was part of the initial projectile) is 255 m + 1020 m = 1275 m.

 

I hope this helps!  Just let me know if you have any questions about this, or catch any errors (which I am known to make :) ).