A projectile fired from a cannon mounted at an angle of $45^{\circ}$ to the horizon on a flat horizontal plain breaks at the upper point of its trajectory into two fragments of equal mass. The first fragment falls to the ground directly below the point of bursting 15 s later. At what distance from the cannon will the second fragment fall if 15 s elapsed from the moment of firing to the moment of rupture? Neglect air resistance.
Decision:
Since the time of fall of the first fragment and the time of ascent of the projectile to the upper point of the trajectory (where the vertical component of its velocity goes to zero) are equal, the initial velocity of the first fragment is zero.
Let $l_{1}$ be the distance from the gun to the place where the first fragment falls, $l_{2}$ be the distance from the place where the first fragment falls to the place where the second fragment falls, then the required distance from the gun to the place where the second fragment falls $l$ will be equal to:
$l=l_{1}+l_{2}$.
Let us decompose the initial velocity of the projectile $\bar{v_{0}}$ into two components: $v_{0x}$ - along the x-axis directed horizontally in the direction of the projectile movement, and $v_{0y}$ - along the y-axis directed vertically upwards.
Since the angle is $45^{ \circ }$, then $v_{0x} = v_{0y}$; $v_{0y}$ is found from the condition that the vertical component of velocity at the upper point of the trajectory is zero:
$v_{0y}=gt_{1}$,
where $t_{1}$ is the time from the moment of firing to the moment of projectile burst. Then it follows from the equality $l_{1}=v_{0x}t_{1}$ that $l_{1}=gt^{2}_{1}$. From the law of conservation of the horizontal component of momentum
$ m v_{0x}= \frac{m}{2} v_{2} v_{2x}$
we find the horizontal component of the velocity of the second fragment $v_{2x}=2v_{0x}$. It follows from the law of conservation of the vertical component of momentum that the vertical component of the velocity of the second fragment is zero. It falls during the same time $t_{1}$ as the first fragment and travels horizontally a path $l_{2}=v_{2x}t_{1} \: \: 2v_{0x}t_{1}=2gt^{2}_{1}$.
So, $l = l_{1}+l_{2}=3gt^{2}=6615 m$.
Decision:
Since the time of fall of the first fragment and the time of ascent of the projectile to the upper point of the trajectory (where the vertical component of its velocity goes to zero) are equal, the initial velocity of the first fragment is zero.
Let $l_{1}$ be the distance from the gun to the place where the first fragment falls, $l_{2}$ be the distance from the place where the first fragment falls to the place where the second fragment falls, then the required distance from the gun to the place where the second fragment falls $l$ will be equal to:
$l=l_{1}+l_{2}$.
Let us decompose the initial velocity of the projectile $\bar{v_{0}}$ into two components: $v_{0x}$ - along the x-axis directed horizontally in the direction of the projectile movement, and $v_{0y}$ - along the y-axis directed vertically upwards.
Since the angle is $45^{ \circ }$, then $v_{0x} = v_{0y}$; $v_{0y}$ is found from the condition that the vertical component of velocity at the upper point of the trajectory is zero:
$v_{0y}=gt_{1}$,
where $t_{1}$ is the time from the moment of firing to the moment of projectile burst. Then it follows from the equality $l_{1}=v_{0x}t_{1}$ that $l_{1}=gt^{2}_{1}$. From the law of conservation of the horizontal component of momentum
$ m v_{0x}= \frac{m}{2} v_{2} v_{2x}$
we find the horizontal component of the velocity of the second fragment $v_{2x}=2v_{0x}$. It follows from the law of conservation of the vertical component of momentum that the vertical component of the velocity of the second fragment is zero. It falls during the same time $t_{1}$ as the first fragment and travels horizontally a path $l_{2}=v_{2x}t_{1} \: \: 2v_{0x}t_{1}=2gt^{2}_{1}$.
So, $l = l_{1}+l_{2}=3gt^{2}=6615 m$.