Question

The distance covered by a particle thrown in a vertical plane in horizontal and vertical directions at any instance of time $t$ are $x=\sqrt{21}tm$ and $y=(2t-4t^2)m$ respectively. The initial velocity of the particle (in $m/s$) is

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is A $5$

Second equation of motion in horizontal and vertical direction of motion,
$x\text{- direction}:x=u_{x}t$
$y\text{- direction}:y=u_{y}t–\frac{1}{2}g{t}^2$
Compare the above equations from the equations given in questions, we get $u_x=\sqrt{21};u_y=2$
Hence initial velocity of particle is the resultant of these two components of velocity
$u=\sqrt{u_x^2+u_y^2}$
$=\sqrt{21+4}$
$=5m{s}^{-1}$