Question

When a body is projected with velocity $\overrightarrow{u}$ horizontally from a height $h$.

Column-I	Column-II
(a)	The speed of body at any time $t$ is	(p)	$\sqrt{\frac{2gh}{u^2}}$
(b)	It will strike the ground after time $t$ equal to	(q)	$\sqrt{\frac{2h}{g}}$
(c)	The path followed by the body is expressed as $y$ equal to	(r)	$-\frac{1}{2}g\frac{x^2}{u^2}$
(d)	The tangent of the angle $\theta$ made by velocity vector striking the ground with horizontal is equal to	(s)	$\sqrt{u^2+g^2{t}^2}$

• A. a-s, b-p, c-q, d-q
• B. a-s, b-p, c-s, d-p
• C. a-s, b-q, c-r, d-p
• D. a-r, b-p, c-q, d-q

Answer 1

The correct option is C a-s, b-q, c-r, d-p

$\left(a\right)\to \left(s\right);\left(b\right)\to \left(q\right);\left(c\right)\to \left(r\right);\left(d\right)\to \left(p\right)$
(a) $v=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+{\left(gt\right)}^2}=\sqrt{u^2+g^2{t}^2}$
$\left(b\right)att=T,y=0$
$\therefore 0=h-\frac{1}{2}g{t}^2$
$\therefore t=\sqrt{\frac{2h}{g}}$
$\left(c\right)y=-\frac{1}{2}\frac{g{x}^2}{u^2}$
$\left(d\right)\frac{dy}{dx}=\frac{-gx}{u^2}$
at t = T(when it strikes the ground)
x = R