Question: The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain plan...

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

Answer 1

Solution

$v_{y} = \frac{dy}{dt} = 8 - 10t$ , $v_{x} = \frac{dx}{dt} = 6$

at the time of projection i.e. $v_{y} = \frac{dy}{dt} = 8$ and $v_{x} = 6$

$\therefore v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10\mspace{6mu} m ⥂ / ⥂ s$

Question: The height \(y\) and the distance \(x\) along the horizontal plane of a projectile on a certain plan...

Solution