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3-2.Motion in Plane
medium
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 acceleration due to gravity is given by ......... $m/{\sec ^2}$
A
$10$
B
$5$
C
$20$
D
$2.5$
Solution
(a) ${a_x} = \frac{d}{{dt}}({v_x}) = 0$,
${a_y} = \frac{d}{{dt}}({v_y}) = – 10\;m/{s^2}$
Net acceleration $a = \sqrt {a_x^2 + a_y^2} $=$\sqrt {{0^2} + {{10}^2}} =10m/s^2$
Standard 11
Physics
