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# A body projected horizontally with a velocity $v$ from a height $h$ has a range $R$. With what velocity this body has to be projected horizontally from a height $\dfrac{h}{2}$ to have the same range?A. $\sqrt 2 v$B. $2v$C. $6\,v$D. $8\,v$

Last updated date: 14th Jun 2024
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Hint: We can use the equation of motion for vertical displacement and horizontal displacement to find the answer. In the vertical displacement there is acceleration due to gravity. Whereas in the horizontal direction there is no acceleration due to gravity. Using this we can find the equation for range and height for both the cases. Then by comparing the equations we can find the final velocity with which it should be projected for getting the same range.

It is given that a body is projected horizontally from a height h with a velocity v.
The range is R.
We need to find the velocity of this body when it is projected from a height $\dfrac{h}{2}$ such that it has the same range.
Consider the figure given below

Height h is the vertical displacement.
We know the second equation of motion given as
$s = ut + \dfrac{1}{2}a{t^2}$
Where, s is the displacement, u is the initial velocity, t is the time taken, a is the acceleration.
So, the equation for the vertical displacement can be written as
$h = 0 \times t + \dfrac{1}{2}g{t^2}$
Here the velocity in y direction can be taken zero initially because the body is projected horizontally so there is only velocity in x direction.
$\Rightarrow h = \dfrac{1}{2}g{t^2}$ (1)
Now let us write the equation of motion for the horizontal displacement which is the range R .
The acceleration due to gravity acts only in the vertical direction hence in this case g can be taken as zero.
$\Rightarrow R = vt + 0$
$\Rightarrow R = vt$
$\Rightarrow t = \dfrac{R}{v}$ (2)
Now substituting the value of t from equation to in equation 1 we get
$\Rightarrow h = \dfrac{1}{2}g{\left( {\dfrac{R}{v}} \right)^2}$ (3)
Now let us analyze the second case. Let the velocity with which it is projected is $v'$ .

The equation for vertical displacement can be written as
$\dfrac{h}{2} = 0 \times t' + \dfrac{1}{2}g{t'^2}$
$\Rightarrow \dfrac{h}{2} = \dfrac{1}{2}g{t'^2}$ (4)
The equation for the horizontal displacement can be written as
$R = v't' + 0$
$\Rightarrow R = v't'$
$\Rightarrow t' = \dfrac{R}{{v'}}$ (5)
Substituting the value of $t'$ in equation 4 we get
$\Rightarrow \dfrac{h}{2} = \dfrac{1}{2}g{\left( {\dfrac{R}{{v'}}} \right)^2}$ (6)
Now let us divide equation 3 by equation 6.
Then we get
$\Rightarrow \dfrac{{2h}}{h} = \dfrac{{\dfrac{1}{2}g{{\left( {\dfrac{R}{v}} \right)}^2}}}{{\dfrac{1}{2}g{{\left( {\dfrac{R}{{v'}}} \right)}^2}}}$
$\Rightarrow 2 = {\left( {\dfrac{{v'}}{v}} \right)^2}$
$\Rightarrow \sqrt 2 = \dfrac{{v'}}{v}$
$\therefore v' = v\sqrt 2$
This is the value of velocity with which it should be projected for getting the same range $R$ at a height $\dfrac{h}{2}$.
So, the correct answer is option A.

Note: Always remember that gravity will accelerate a body only in the vertical direction. Gravity will not pull a projectile sideways. It will only pull in the downward direction. So, while writing the equation of motion for the horizontal displacement the acceleration due to gravity should be taken as zero.