Question

A projectile is fixed with certain velocity $u$ making an angle $\theta$ with the horizontal. With necessary diagram prove that the trajectory of the projectile is a parabola and also obtain an expression for

(i)Time of flight

(ii)max height reach

(iii)horizontal range of projectile

Answer 1

Consider the following equations of a projectile with angle

of projection $\theta$ and initial velocity $v_0$ - here

$x=v_{ox}t$ refer to book ;

rearrange the expression for time t ,

$t=\frac{x}{v_{0x}}$

substituted $\frac{x}{v_{ox}}$ for in the expression

$y=\left(v_{0}sin\theta \right)t-\frac{1}{2}g{t}^2$

$y=v_{oy}\left(\frac{x}{v_{ox}}\right)-\frac{1}{2}g(\frac{x}{v_{0x}}{)}^2$

substitute $v_{o}cos\theta$ for $v_{ox}sin$ $\theta$ for $v_{oy}$

in the above expression.

$Y=\left(v_{0}sin\theta \right)\left(\frac{x}{v_{o}cos\theta }\right)-\frac{1}{2}g(\frac{x}{v_{o}cos\theta }{)}^2$

Hence the obtained equation of the projectile is

$Y=\left(tan\theta \right)x-\left(\frac{gse{c}^2\theta }{2(v_o{)}^2}\right)x^2$

The expression is to be obtained in the form of

Y = $ax+b{x}^2$ .

hence it is a parabolic motion so from it we get all the required quantities

( 1 ) T = $\frac{2v_{o}sin\theta }{g}$

( 2 ) $H_{max}=\frac{v_o^{2}si{n}^2\theta }{2g}$

( 3 ) Range = $\frac{v_o^{2}sin2\theta }{g}$