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 $y=(8t-5t^2)$ 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. $v_y=\frac{dy}{dt}=8and{v}_x=6$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{6^2+8^2}=10m/s$