The height $y$ and the distance $x$ along the vertical plane of a projectile on a certain planet (with no surr

Question

The height $y$ and the distance $x$ along the vertical 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:

Accepted Answer

1.  **Velocity Components:** The velocity of a particle is defined as the time rate of change of its position. The components of velocity are given by $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$ .
2.  **Differentiate Position Equations:**
    *   Given $x = 6t$, the horizontal velocity is $v_x = \frac{d}{dt}(6t) = 6 	ext{ m/s}$.
    *   Given $y = 8t - 5t^2$, the vertical velocity is $v_y = \frac{d}{dt}(8t - 5t^2) = 8 - 10t 	ext{ m/s}$.
3.  **Initial Velocity (at t=0):** The projectile is projected at time $t = 0$. Substituting $t=0$ into the velocity components:
    *   $v_{x0} = 6 	ext{ m/s}$.
    *   $v_{y0} = 8 - 10(0) = 8 	ext{ m/s}$.
4.  **Calculate Resultant Velocity:** The magnitude of the initial velocity (speed of projection) is:
    $v = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 	ext{ m/s}$ .

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