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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 angle with the horizontal at which the projectile was projected is
A
${\tan ^{ - 1}}(3/4$)
B
${\tan ^{ - 1}}(4/3)$
C
${\sin ^{ - 1}}(3/4$)
D
Not obtainable from the given data
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 ${v_x} = 6$
The angle of projection is given by $\theta = {\tan ^{ – 1}}\left( {\frac{{{v_y}}}{{{v_x}}}} \right) $
$= {\tan ^{ – 1}}\left( {\frac{4}{3}} \right)$
Standard 11
Physics
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