Answer
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Hint: In this question, we will use the relation of the horizontal range R, time T to find the speed S of the shot. This will give us a final relation between range and speed, using this relation we will deduce the required expression and get the result. Further, we will discuss the basics of time of flight and how height changes the energy of an object.
Formula used:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$
$T = \dfrac{{2v\sin \phi }}{g}$
Complete answer:
Here, we will use the relation of the horizontal range R, time T to find the speed of the shot, which is given as
As we know, the horizontal range R is given by:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$
Time of flight T is given by:
$T = \dfrac{{2v\sin \phi }}{g}$
So, the distance moved by P in the given time T is given by:
$S = \dfrac{1}{2}a{\left( {\dfrac{{2v\sin \phi }}{g}} \right)^2}$
Therefore, the changed range is given by the addition of horizontal range R and distance moved by P in given time T, this can be written as:
$\dfrac{{{v^2}\sin 2\phi }}{g} = \dfrac{{{v^2}\sin 2\theta }}{g} + \dfrac{{2a{v^2}{{\sin }^2}\phi }}{{{g^2}}}$
$ \Rightarrow g\sin 2\phi = g\sin 2\theta + 2a{\sin ^2}\phi $
Now, we solve for R.H.S:
$g\sin \theta + 2a\left( {\dfrac{{1 - \cos 2\phi }}{2}} \right)$
$ = g\sin 2\theta + a - a\cos 2\phi $
Now, again when we put this value of R.H.S in above equation we get:
$ \Rightarrow g\sin 2\phi - g\sin 2\theta = a - a\cos 2\phi $
Solving for n, we get:
$ \Rightarrow g(\sin 2\phi - \sin 2\theta ) = a(1 - \cos 2\phi )$
$\eqalign{& \Rightarrow \dfrac{g}{a} = \dfrac{{1 - \cos 2\phi }}{{\sin 2\phi - \sin 2\theta }} \cr
& \therefore n = 2 \cr} $
Therefore, we get the required result, i.e., n=2, which is given by the above value.
Additional information:
As we know that Time of flight is defined as the measurement of the time taken by an object or particle to travel a distance through a particular medium.
Also, when an object or particle is at a certain height, it gains potential energy.
Further, as we know that energy is defined as the ability to do work. Energy can be found in many things and can take different forms, like-kinetic energy is the energy of motion, and potential energy is energy due to an object's position or structure.
We also know that, according to the law of conservation of energy, energy can neither be created nor be destroyed; rather it can only be transferred from one form to another. Here, in this question electrical energy is transferred to heat energy.
We also know the basic difference of speed and velocity i.e., speed is the measure of how fast an object can travel, whereas velocity tells us the direction of this speed. Speed is a scalar quantity that means it has only magnitude, whereas velocity is a vector quantity that means it has both magnitude and direction. The S.I unit of velocity is meter per second (m/sec).
Note:
Here, we should remember that time of flight can be calculated by wave as well. Also, total energy is given by the sum of potential energy and kinetic energy. The unit of energy here is electron volt which is $1.602 \times {10^{ - 19}}Joule$ . The S.I unit of energy is Joule. At the maximum height the potential energy is maximum and kinetic energy is zero
Formula used:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$
$T = \dfrac{{2v\sin \phi }}{g}$
Complete answer:
Here, we will use the relation of the horizontal range R, time T to find the speed of the shot, which is given as
As we know, the horizontal range R is given by:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$
Time of flight T is given by:
$T = \dfrac{{2v\sin \phi }}{g}$
So, the distance moved by P in the given time T is given by:
$S = \dfrac{1}{2}a{\left( {\dfrac{{2v\sin \phi }}{g}} \right)^2}$
Therefore, the changed range is given by the addition of horizontal range R and distance moved by P in given time T, this can be written as:
$\dfrac{{{v^2}\sin 2\phi }}{g} = \dfrac{{{v^2}\sin 2\theta }}{g} + \dfrac{{2a{v^2}{{\sin }^2}\phi }}{{{g^2}}}$
$ \Rightarrow g\sin 2\phi = g\sin 2\theta + 2a{\sin ^2}\phi $
Now, we solve for R.H.S:
$g\sin \theta + 2a\left( {\dfrac{{1 - \cos 2\phi }}{2}} \right)$
$ = g\sin 2\theta + a - a\cos 2\phi $
Now, again when we put this value of R.H.S in above equation we get:
$ \Rightarrow g\sin 2\phi - g\sin 2\theta = a - a\cos 2\phi $
Solving for n, we get:
$ \Rightarrow g(\sin 2\phi - \sin 2\theta ) = a(1 - \cos 2\phi )$
$\eqalign{& \Rightarrow \dfrac{g}{a} = \dfrac{{1 - \cos 2\phi }}{{\sin 2\phi - \sin 2\theta }} \cr
& \therefore n = 2 \cr} $
Therefore, we get the required result, i.e., n=2, which is given by the above value.
Additional information:
As we know that Time of flight is defined as the measurement of the time taken by an object or particle to travel a distance through a particular medium.
Also, when an object or particle is at a certain height, it gains potential energy.
Further, as we know that energy is defined as the ability to do work. Energy can be found in many things and can take different forms, like-kinetic energy is the energy of motion, and potential energy is energy due to an object's position or structure.
We also know that, according to the law of conservation of energy, energy can neither be created nor be destroyed; rather it can only be transferred from one form to another. Here, in this question electrical energy is transferred to heat energy.
We also know the basic difference of speed and velocity i.e., speed is the measure of how fast an object can travel, whereas velocity tells us the direction of this speed. Speed is a scalar quantity that means it has only magnitude, whereas velocity is a vector quantity that means it has both magnitude and direction. The S.I unit of velocity is meter per second (m/sec).
Note:
Here, we should remember that time of flight can be calculated by wave as well. Also, total energy is given by the sum of potential energy and kinetic energy. The unit of energy here is electron volt which is $1.602 \times {10^{ - 19}}Joule$ . The S.I unit of energy is Joule. At the maximum height the potential energy is maximum and kinetic energy is zero
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