Height y and distance x along horizontal plane of a projectile on a certain planet (with no surround

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3-2.Motion in Plane

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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 velocity with which the projectile is projected is ......... $m/sec$.

A

$8$

B

$6$

C

$10$

D

Not obtainable from the data

Solution

(c) ${v_y} = \frac{{dy}}{{dt}} = 8 – 10t$, ${v_x} = \frac{{dx}}{{dt}} = 6$

at the time of projection i.e. ${v_y} = \frac{{dy}}{{dt}} = 8$and ${v_x} = 6$

$\therefore v = \sqrt {v_x^2 + v_y^2} = \sqrt {{6^2} + {8^2}} = 10\;m/s$

Standard 11

Physics

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