ux, ax and uy, ay
where $\text{u}_\text{x}=\text{u}\cos\theta,\text{ u}_\text{y}=\text{u}\sin\theta,$
ax = 0, ay = -g g is the acceleration due to gravity.
The coordinates of O are (0, 0) considering horizontal motion. The position of the body after time t has coordinate (x, y); where $\text{x}(\text{t})=\text{x}_0+\text{u}_\text{x}\text{t}+\frac{1}{2}\text{a}_\text{x}\text{t}^2$ Substituting for various factors $\text{x (t)}=0+\text{u}\cos\theta.\text{t}+\frac{1}{2}\times0\times\text{t}^2$ or $\text{x (t)}=\text{u}\cos\theta.\text{t}$ $\text{t}=\frac{\text{x (t)}}{\text{u}\cos\theta}\dots(\text{i})$
Considering the vertical motion $\text{y (t)}=\text{y}(0)+\text{u}_\text{y}+\frac{1}{2}\text{a}_\text{y}\text{t}^2$ or $\text{y (t)}=0+\text{u}\sin\theta.\text{t}-\frac{1}{2}\text{gt}^2$ or $\text{y (t)}=\text{u}\sin\theta.\text{t}-\frac{1}{2}\text{gt}^2\dots(\text{ii})$ Substituting for t from equation (i) in equation (ii), we get $\text{y (t)}=\text{u}\sin\theta\Big(\frac{\text{x (t)}}{\text{u}\cos\theta}\Big)-\frac{1}{2}\text{g}\Big(\frac{\text{x (t)}}{\text{u}\cos\theta}\Big)^2\text{s}$ $\Rightarrow\ \text{y (t)}=\text{x (t)}\tan\theta-\frac{1}{2}\text{g}\Big(\frac{\text{x}^2(\text{t})}{\text{u}^2\cos^2\theta}\Big)\dots(\text{iii})$ This is an equation of parabola. Thus, the path of a projectile is a parabola. Maximum height attained: At the maximum height, the vertical component of velocity becomes zero. Now using the equation of motion. $\text{h}=\frac{\text{v}^2_\text{y}-\text{u}^2_\text{y}}{2\text{a}_\text{y}}$ We have maximum height$\therefore\ \text{h}_\text{max}=\frac{0^2-(\text{u}\sin\theta)^2}{2(-\text{g})}$
or
$\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}\dots(\text{iv})$ Time for maximum height: Using equation of motion v = u + at or vx = ux + ayt we have $0=\text{u}\sin\theta-\text{gt}$ or $\text{t}=\frac{\text{u}\sin\theta}{\text{g}}\dots(\text{v})$ Horizontal range: Let the horizontal range be x. Since there is no acceleration in the horizontal direction so $\text{x}=\text{x}(0)+\text{u}_\text{x}\text{t}+\frac{1}{2}\text{a}_\text{x}\text{t}^2$As x(0) = 0,
$\text{u}_\text{x}=\text{u}\cos\theta,$ ax = 0 and it is the total time of the flight which is twice the time for maximum height because body takes same time in rising to and falling from the highest point.Hence,
$\text{t}=\frac{2\text{u}\sin\theta}{\text{g}}$$\therefore \ \text{x}=0+\text{u}\cos\theta.\text{t}$
$=\text{u}\cos\theta\Big(\frac{2\text{u}\sin\theta}{\text{g}}\Big)$
or
$\text{x}=\frac{\text{u}^2}{\text{g}}(2\sin\theta\cos\theta)$$\Rightarrow\ \text{x}=\frac{\text{u}^2}{\text{g}}\sin2\theta\dots(\text{vi})$
Maximum horizontal range: From equation (vi) for x to be maximum, the value of $\sin2\theta$ should be maximum which is 1, Hence, $\text{x}_\text{max}=\frac{\text{u}^2}{\text{g}}\dots(\text{vii})$ For this xmax, $\sin2\theta=1\Rightarrow\ \theta=45^\circ$ Therefore, the horizontal range will be maximum if the angle of projection is 45° or $\frac{\pi}{4}$ radians. Time of flight of the projectiles: The projectile after completing its flight returns back to the same horizontal level from which it was projected. Therefore, the vertical displacement in the whole flight is zero. Considering vertical motion. $\text{y (t)}=\text{y}(0)+\text{u}_\text{y}\text{t}+\frac{1}{2}\text{a}_\text{y}\text{t}^2$ Now, y(t) = 0, y(0) = 0, $\text{u}_\text{y}=\text{u}\sin\theta,$ ay = -g Then $0=0+\text{u}\sin\theta.\text{T}-\frac{1}{2}\text{gT}^2$ $\Rightarrow\ \text{T}\Big(\text{u}\sin\theta-\frac{1}{2}\text{gT}\Big)\text{s}=0$ Therefore T = 0 and $\text{u}\sin\theta-\frac{1}{2}\text{gT}=0$ $\Rightarrow\ \text{u}\sin\theta=\frac{1}{2}\text{gT}$ or $\text{gT}=2\text{u}\sin\theta$ $\text{T}=\frac{2\text{u}\sin\theta}{\text{g}}\dots(\text{viii})$ Equation (viii) gives the total time of flight. This is twice the time for maximum height.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
