Question

For an object projected from the ground with speed u, horizontal range is two times the maximum height attained by it. The horizontal range of the object is:

Answer 1

Hint: Here, we will use the expression for the horizontal range and the maximum height attained by the object. This will help us to the required result. Further, we will discuss the effects of maximum height on potential and kinetic energy. Also, we will see the basic difference between the speed and the velocity.

Formula used:
$\eqalign{
  & R = \dfrac{{{u^2}\sin 2\theta }}{g} \cr
  & H = 2 \times \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \cr} $

Complete step by step solution:
Here, we will use the relation between the horizontal range R, acceleration due to gravity g, initial speed u and the maximum height H, which is given as:
$\eqalign{& R = H \cr
  & \Rightarrow \dfrac{{{u^2}\sin 2\theta }}{g} = 2 \times \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \cr} $
$\eqalign{
  & \Rightarrow 2\sin \theta \cos \theta = {\sin ^2}\theta \cr
  & \Rightarrow \tan \theta = 2 \cr} $
From this we get the following values:
$\sin \theta = \dfrac{2}{{\sqrt 5 }}$
$\cos \theta = \dfrac{1}{{\sqrt 5 }}$
Now, by substituting these values to find the horizontal range R:
$\eqalign{
  & R = \dfrac{{{u^2} \times 2\sin \theta \cos \theta }}{g} \cr
  & \Rightarrow R = \dfrac{{{u^2} \times 2 \times \dfrac{2}{{\sqrt 5 }} \times \dfrac{1}{{\sqrt 5 }}}}{g} \cr
  & \therefore R = \dfrac{{4{u^2}}}{{5g}} \cr} $

Therefore, we get the required horizontal range R of the object, projected from the ground with speed u.

Additional information:
As we know that 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, 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.