Question

If the horizontal range of the projectile is\[a\] and maximum height is \[b\] then prove that the velocity of the projectile is ​ \[{\left[ {2g\{ b + \dfrac{{{a_2}}}{{16}}\} } \right]^{1/2}}\]

Answer 1

Hint: Projectile motion is a motion that is experienced by an object that is thrown in the air and moves along the curved path under the action of gravity. When an object is thrown horizontally the object covers a maximum distance. When an object is thrown vertically it covers a maximum height. The horizontal range is defined as the horizontal distance traveled by the body during the time of flight.

Complete answer:
The mathematical expression for the horizontal range of the projectile motion \[R\] is given by,
\[R = \dfrac{{{u^2}\sin 2\theta }}{g}\]
Here \[u\] is the velocity with which the object travels.
\[\theta \] is the projectile angle.
\[g\] is the acceleration due to gravity.
The maximum horizontal range will be acquired when \[\sin \theta \]\[ = 4{5^0}\]
The mathematical expression for the maximum height of the projectile motion\[h\]is given by,
\[h = \dfrac{{{u^2} \cdot si{n^2}\theta }}{{2 \cdot g}}\]
Given in the question, the horizontal range of the projectile is \[a\] and maximum height is \[b\].
Therefore,
\[a = \dfrac{{{u^2}\sin 2\theta }}{g}\] …… (1)
\[b = \dfrac{{{u^2} \cdot si{n^2}\theta }}{{2 \cdot g}}\] …… (2)
From equation (1), rearranging this equation we get,
\[\sin 2\theta = \dfrac{{ag}}{{{u^2}}}\]
Squaring on both sides,
\[{(\sin 2\theta )^2} = \dfrac{{{a^2}{g^2}}}{{{u^4}}}\] …… (3)
From equation (2), rearranging this equation we get,
\[si{n^2}\theta = \dfrac{{2bg}}{{{u^2}}}\]
\[1 - 2si{n^2}\theta = 1 - \dfrac{{4bg}}{{{u^2}}}\]
From the trigonometric formula,
\[\cos 2\theta = 1 - 2si{n^2}\theta \]
Therefore the above equation becomes,
\[\cos 2\theta = 1 - \dfrac{{4bg}}{{{u^2}}}\]
Squaring on both sides,
\[{(\cos 2\theta )^2} = {(1 - \dfrac{{4bg}}{{{u^2}}})^2}\] ……. (4)
Adding (3) and (4)
\[{\sin ^2}2\theta + {\cos ^2}2\theta = \dfrac{{{a^2}{g^2}}}{{{u^4}}} + {(\dfrac{{{u^2} - 4bg}}{{{u^2}}})^2}\]
We know that, \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Therefore the above equation becomes,
\[1 = \dfrac{{{a^2}{g^2}}}{{{u^4}}} + {\dfrac{{({u^2} - 4bg)}}{{{u^4}}}^2}\]
Solving for \[u\]we get,
\[u = {\left[ {2g\{ b + \dfrac{{{a_2}}}{{16}}\} } \right]^{1/2}}\]
Hence Proved.

Note:
The body with projectile motion starts with a velocity. This velocity will be resolved into the horizontal and vertical components. The horizontal component of the velocity will not be acted upon by the force. But for the vertical component, a downward acceleration will be acting on the body. Therefore the horizontal component of the velocity will remain unchanged throughout the projectile motion. But due to the gravitational force, the vertical component will be changing throughout the projectile motion.