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# Show that, A) For a projectile the angle between the velocity and the X-axis as a function of given by $\theta \left( t \right) = {\tan ^{ - 1}}\left( {\dfrac{{{{\left( {{v_0}} \right)}_y} - gt}}{{{{\left( {{v_0}} \right)}_x}}}} \right)$(B) Show that the projection angle ${\theta _0}$ for a projectile launched from the origin is given by ${\theta _0} = {\tan ^{ - 1}}\left( {\dfrac{{4{h_m}}}{R}} \right)$where the symbols have their usual meaning.

Last updated date: 22nd Feb 2024
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Hint: To solve part (A) we have to use the relation between angle of velocity of particle after time $t$ when object reached at a point on projectile path and to find the initial angle we calculate time in which object reached at the highest point on the path and find then we can find the relation between ${h_m}$,$R$ and angle of projection.

Complete step by step solution:
We assume an object of mass $m$ projected with initial velocity ${v_0}$ at an angle ${\theta _0}$ with the horizontal and after time $t$ it reaches at another point on projectile path where velocity of object become $v$ and angle $\theta$ with the horizontal as shown in figure

Initial velocity is ${v_0}$ its component
Horizontal component ${\left( {{v_0}} \right)_x} = {v_0}\cos {\theta _0}$ ................ (1)
Vertical component ${\left( {{v_0}} \right)_y} = {v_0}\sin {\theta _0}$ .................. (2)
And its velocity after time $t$ become $v$ then its components
Horizontal component ${v_x} = v\cos \theta$ ................ (3)
Vertical component ${v_y} = v\sin \theta$ .................... (4)
Part (A)
For vertical motion of object we use equation of motion
$\Rightarrow {v_y} = {\left( {{v_0}} \right)_y} - gt$ ................... (5)
And we know the horizontal component of velocity in projectile motion remain same so the initial and final horizontal component are equal
$\Rightarrow {v_x} = {\left( {{v_0}} \right)_x}$........... (6)
(4) Divided by (3)
$\Rightarrow \dfrac{{v\sin \theta }}{{v\cos \theta }} = \dfrac{{{v_y}}}{{{v_x}}}$
$\Rightarrow \tan \theta = \left( {\dfrac{{{v_y}}}{{{v_x}}}} \right)$
From equation (5)
$\Rightarrow \tan \theta = \left( {\dfrac{{{{\left( {{v_0}} \right)}_y} - gt}}{{{v_x}}}} \right)$
From equation (6) ${v_x} = {\left( {{v_0}} \right)_x}$
$\Rightarrow \tan \theta = \left( {\dfrac{{{{\left( {{v_0}} \right)}_y} - gt}}{{{{\left( {{v_0}} \right)}_x}}}} \right)$

Part (B)
We assume the maximum height of projectile path is ${h_m}$ where object reached in time $t$ then from first equation of motion for vertical motion
$\Rightarrow {v_y} = {\left( {{v_0}} \right)_y} - gt$
Vertical component of velocity at maximum height will be zero so
$\Rightarrow 0 = {v_0}\sin {\theta _0} - gt$
So time to reach at maximum height is
$\Rightarrow t = \dfrac{{{v_0}\sin {\theta _0}}}{g}$ ........... (7)
Now from second equation of motion
$\Rightarrow {h_m} = {\left( {{v_0}} \right)_y}t - \dfrac{1}{2}g{t^2}$
$\Rightarrow {h_m} = {v_0}\sin {\theta _0} \times \dfrac{{{v_0}\sin {\theta _0}}}{g} - \dfrac{1}{2}g \times {\left( {\dfrac{{{v_0}\sin {\theta _0}}}{g}} \right)^2}$
Solving this
$\Rightarrow {h_m} = \dfrac{{{v_0}^2{{\sin }^2}{\theta _0}}}{{2g}}$ ............ (8)
For horizontal motion
$\Rightarrow R = {\left( {{v_0}} \right)_x} \times T$
Put the value of flight time $T = 2t = \dfrac{{2{v_0}\cos {\theta _0}}}{g}$
$\Rightarrow R = {v_0}\cos {\theta _0} \times \dfrac{{2{v_0}\sin {\theta _0}}}{g}$
$\Rightarrow R = \dfrac{{2{v_0}^2\sin {\theta _0}\cos {\theta _0}}}{g}$ ................... (9)
Equation (8) divides by (9)
$\Rightarrow \dfrac{{{h_m}}}{R} = \dfrac{{\dfrac{{{v_0}^2{{\sin }^2}{\theta _0}}}{{2g}}}}{{\dfrac{{2{v_0}^2\sin {\theta _0}\cos {\theta _0}}}{g}}}$
$\Rightarrow \dfrac{{{h_m}}}{R} = \dfrac{{{v_0}^2{{\sin }^2}{\theta _0}}}{{2g}} \times \dfrac{g}{{2{v_0}^2\sin {\theta _0}\cos {\theta _0}}}$
Farther solving
$\Rightarrow \dfrac{{{h_m}}}{R} = \dfrac{{\sin {\theta _0}}}{{4\cos {\theta _0}}}$
$\Rightarrow \dfrac{{4{h_m}}}{R} = \dfrac{{\sin {\theta _0}}}{{\cos {\theta _0}}}$
We can write

$\therefore \tan {\theta _0} = \dfrac{{4{h_m}}}{R}.$

Note: When an object move in projectile motion the horizontal component of velocity never changes because there is no acceleration in horizontal direction so it will remain same in whole motion and the vertical component of velocity changes due to the gravitational acceleration when the object moving in upward direction then it gradually decreases with height and becomes zero at maximum point of path.