Answer the following in Brief

Question

Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity u at an angel to the horizontal.

Vidyadip · Accepted Answer

Expression for range:
1. Consider a body projected with velocity $\vec{u}$, at an angle $\theta$ of projection from point $O$ in the co-ordinate system of the $X Y$ - plane, as shown in figure.
2. The initial velocity $\vec{u}$ can be resolved into two rectangular components:

$u_x = u \cos \theta $ (Horizontal component)
$u_y = u \sin \theta $ (Vertical component)
The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to
$v_y = u_y + a_yt$
with $a_y = -g$ and $u_y = u \sin \theta $
Thus, the components of velocity of the projectile at time t are given by,
$v_x = u_x = u \cos \theta $

$v_y = u_x – gt = u \sin \theta – gt$

Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
s = (u cosθ) t
$s_y=(u \sin \theta) t -\frac{1}{2} gt ^2$
6. At the highest point, the time of ascent of the projectile is given as, $t _{ A }=\frac{u \sin \theta}{g} \ldots \ldots \ldots \ldots . .(2)$
7. The total time in air i.e., time of flight is given as, $T=2 t_A=\frac{2 u \sin \theta}{g}$
8. The total horizontal distance travelled by the particle in this time $T$ is given as,
$R = u _{ x } \cdot T$
$R = u \cos \theta \cdot\left(2 t _{ A }\right)$
$R = u \cos \theta \cdot \frac{2 u \sin \theta}{ g } \ldots \ldots \ldots \ldots \ldots .[\text { From (3)] }$
$R =\frac{u^2(2 \sin \theta \cdot \cos \theta)}{g}$
$R =\frac{ u ^2 \sin 2 \theta}{ g } \ldots \ldots \ldots \ldots .[\because \sin 2 \theta=2 \sin \cdot \cos \theta]$
This is required expression for horizontal range of the projectile.
Expression for maximum height of a projectile:
The maximum height H reached by the projectile is the distance travelled along the vertical (y) direction in time $t_A.$
Substituting $s_y = H$ and $t = t_a$ in equation (1),
we have,

$H =( u \sin \theta) t _{ A }-\frac{1}{2} gt _{ A }{ }^2$
$\therefore \quad H = u \sin \theta\left(\frac{ u \sin \theta}{ g }\right)-\frac{1}{2} g\left(\frac{ u \sin \theta}{ g }\right)^2$
$\therefore \quad H =\frac{ u ^2 \sin ^2 \theta}{2 g }=\frac{ u _y^2}{2 g }$
....[From (2)]
This equation represents maximum height of projectile.

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