Question

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5$t^2$)
meter and x=6t meter, where t is in second. The velocity with which the projectile is projected is

• A. 8 m/sec
• B. 6 m/sec
• C. 10 m/sec
• D. Not obtainable from the data

Answer 1

The correct option is C 10 m/sec

$v_y$ = $\frac{dy}{dt}$ = 8 - 10t, $v_x$ = $\frac{dx}{dt}$ = 6
at the time of projection i.e. $v_y$ = $\frac{dy}{dt}$ =8 and $v_x$ = 6
$\therefore$ v = $\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}$ = $\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}$ = 10 m/s