Question

A particle is projected (x- z plane) such that its velocity at time t is given by $\overrightarrow{v}=a\tilde{i}+(b-ct)\tilde{j}$ Its range on given by the horizontal plane is given by

• A. $\frac{ba}{c}$
• B. $\frac{2ba}{c}$
• C. $\frac{3ba}{c}$
• D. None

Answer 1

The correct option is B $\frac{2ba}{c}$

$v=a\hat{i}+\left(b-ct\right)\hat{j}$

$vx=ux+axt$

$a=ux+0$

$ux=a$

$x=uxt+0$

$x=at$-----------------$\left(1\right)$

$vy=uy+ayt$

$b-ct=uy-gt$

$uy=b$ and $g=c$

Now$,$

$y=uyt+\frac{1}{2}ay{t}^2$

$0=bt-\frac{1}{2}c{t}^2$

$2b-ct=0$

$t=\frac{2b}{c}$-------------$\left(2\right)$

put $\left(2\right)$ in equation $\left(1\right)$

$x=\frac{a2b}{c}=\frac{2ab}{c}$

Hence,

option $\left(B\right)$ is correct answer.