Question

The maximum horizontal range of a projectile is 400m.The maximum value of height attained by it will be?

Answer 1

Let us consider a body, which is projected at an angle,$\theta$ . The maximum horizontal range is given and thus we can find the value of ‘$\theta$’ .Then, the maximum value of height is obtained by the suitable formula.

Formula Used:
${\text{Horizontal Range}}\left( R \right){\text{ = }}\dfrac{{{u^2}\sin 2\theta }}{g}$
${\text{Vertical Height}}\left( H \right){\text{ = }}\dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Where, u is initial velocity in m/sec.
$\theta$ is the angle of projection of the projectile.
g is acceleration due to gravity.

Complete step by step answer:
When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
Along the x-axis: uniform velocity (u cos$\theta$), responsible for the horizontal (forward) motion of the particle.
Along the y-axis: uniform acceleration (g =9.8 m/s$^2$), responsible for the vertical (downwards) motion of the particle.
“Image drawn by SME”

As, we know that horizontal range is given by the formula:
${\text{Horizontal Range}}\left( R \right){\text{ = }}\dfrac{{{u^2}\sin 2\theta }}{g}$
Here, g is fixed i.e., g=9.8 m/s$^2$
u: initial velocity (it will be given in question)
So, the maximum value of range will depend upon sin2$\theta$.
Range will be maximum when ‘sin2$\theta$’ will be maximum.

As, we know that the maximum value of sin2$\theta$is 1.
$\Rightarrow \sin 2\theta = 1 \\\ \Rightarrow \sin 2\theta = \sin {90^ \circ } \\\ \Rightarrow 2\theta = {90^ \circ } \\\ \therefore \theta = {45^ \circ } \\\$
Thus, the range will be maximum when $\theta$=${45^ \circ }$.
${\text{Rang}}{{\text{e}}_{\max }} = \dfrac{{{u^2}}}{g}$ (as,$\sin 2\theta = 1$)

Given, ${\text{Rang}}{{\text{e}}_{\max }} = \dfrac{{{u^2}}}{g}$=400m …………….. (i)
As, we know that vertical height is given by the formula:
${\text{Vertical Height}}\left( H \right){\text{ = }}\dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \\\ \\\$
For, maximum height: $\theta$=${45^ \circ }$.
Putting the value of $\theta$=${45^ \circ }$in vertical height, we get:

{\text{Vertical Height}}\left( H \right){\text{ = }}\dfrac{{{u^2}{{\sin }^2}{{45}^ \circ }}}{{2g}} \\\ \\\ $ Value of sin${45^ \circ }$=$\dfrac{1}{{\sqrt 2 }}$ Putting the value of sin${45^ \circ }$in vertical height, we have: $ {\text{Vertical Height}}{\left( H \right)_{\max }}{\text{ = }}\dfrac{{{u^2}}}{{2g}} \times {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} \\\ \Rightarrow {H_{\max }} = \dfrac{{{u^2}}}{{2g}} \times \dfrac{1}{2} \\\ \therefore {H_{\max }} = \dfrac{{{u^2}}}{{4g}} \\\ \\\ $ Now, putting the value of $\dfrac{{{u^2}}}{g}$ from (i) in $H_{max}$ we have: $ {H_{\max }} = \dfrac{1}{4} \times 400 \\\ \therefore {H_{\max }} = 100m \\\ $ **Hence, the maximum height attained by it will be 100m.** **Note:** Projectile refers to an object that is in flight after being thrown or projected. The path of a projectile is’ parabolic’. Throughout the motion, the acceleration of the projectile is constant and acts vertically downwards being equal to g. There is no acceleration acting in a horizontal direction.