Question

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

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

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

Answer 1

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

Horizontal velocity remains same i.e. $=u$
Vertical velocity at time $t$ after projection is $gt$
$\text{(a)}v=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+(gt{)}^2}=\sqrt{u^2+g^2{t}^2}$
$\text{(b) At}t=T,y=0$
$\therefore 0=h-\frac{1}{2}g{t}^2\therefore t=\sqrt{\frac{2h}{g}}$
$\text{(c)}y=\frac{1}{2}\frac{g{x}^2}{u^2}$
$\text{(d)}\frac{dy}{dx}=\frac{-gx}{u^2}$
at $t=T$ (when it strikes the ground)
$x=R$
$\therefore R=ut=u\times \sqrt{\frac{2h}{g}}$
$\therefore {\left(\frac{dy}{dx}\right)}_{att=T(x=R)}=-\frac{gR}{u^2}$
$\therefore \tan \beta =\frac{-g}{u^2}\times u\sqrt{\frac{2h}{g}}=-\sqrt{\frac{2hg}{u^2}}$
If we consider acute angle then,
$\tan \beta =\sqrt{\frac{2hg}{u^2}}$