Question

The height $y$ and the distance $x$ (both in m) along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=(8t-5t^2)$ metre and $x=6t$ metre where $t$ is in seconds. The velocity of projection is

• A. $8m{s}^{-1}$
• B. $6m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. data insufficient

Answer 1

The correct option is C $10m{s}^{-1}$

To find the resultant velocity let us first find the velocity along the y-axis $v_y$ and $v_x$ along the x-axis.
$v_y=\frac{dy}{dt}=(8-10t)$

$v_x=\frac{dx}{dt}=6$

At $t=0,v_x=6m/s$ and $v_y=8m/s$

$\therefore$ Initially velocity of projection
$=\sqrt{(6{)}^2+(8{)}^2}$
The velocity of projection is $=10m/s$