The height y and the distance x along the hor | vedclass

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(A) The position coordinates of the projectile are given by $x = 6t$ and $y = 8t - 5t^2$.To find the acceleration,we first find the velocity component

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$10$

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(A) The position coordinates of the projectile are given by $x = 6t$ and $y = 8t - 5t^2$.To find the acceleration,we first find the velocity components by differentiating the position with respect to time $t$.Velocity in the $x$-direction: $v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 	ext{ m/s}$.Velocity in the $y$-direction: $v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) = 8 - 10t 	ext{ m/s}$.Now,we find the acceleration components by differentiating the velocity with respect to time $t$.Acceleration in the $x$-direction: $a_x = \frac{dv_x}{dt} = \frac{d}{dt}(6) = 0 	ext{ m/s}^2$.Acceleration in the $y$-direction: $a_y = \frac{dv_y}{dt} = \frac{d}{dt}(8 - 10t) = -10 	ext{ m/s}^2$.The negative sign indicates that the acceleration is acting downwards. The magnitude of the acceleration due to gravity is $g = |a_y| = 10 	ext{ m/s}^2$.

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 $y = (8t - 5t^2) \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The acceleration due to gravity on this planet is ......... $m/s^2$.

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