1. ## help!

A projectile that is fired from a gun has an initial velocity of 90.0 km/h at an angle of 60.0o above the horizontal. When the projectile is at the top of its trajectory, an internal explosion causes it to separate into two fragments of equal mass. One of the fragments falls straight downward as though it had been released from rest. How far from the gun does the other fragment land?

2. Originally Posted by Celia
A projectile that is fired from a gun has an initial velocity of 90.0 km/h at an angle of 60.0o above the horizontal. When the projectile is at the top of its trajectory, an internal explosion causes it to separate into two fragments of equal mass. One of the fragments falls straight downward as though it had been released from rest. How far from the gun does the other fragment land?
The projectile had horizontal velocity(which is constant till the time of explosion) = $90*cos60=45km/h$
I hope that you are familiar with the law of conservation of momentum(which is at the heart of Classical Mechanics).
Total momentum before explosion = Total momentum after explosion
(Let the mass of original projectile be m, then the mass of the fragments will be m/2 and m/2.)
$m*45=\frac{m}{2}*0 + \frac{m}{2}* v$
Solving we get, $v=90km/h$.

I will continue from here(Do you know about maximum height and range?).

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Malay