Question

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 12s after its Firing What is X - and y-coordinates of the rocket relative to its Firing Point?​

Answer 1

688.32m and 277.44m

Step-by-step explanation :

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`\large{\maltese{\textsf{\underline{To find :-}}}}`

The X and Y coordinates of the rocket relative of firing

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`\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^(-1) \\ \sf angle ({\theta}_(1)) = 55.0{\degree} \\ \sf time (t) = 12s`

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`\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}`

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The horizontal range of projectile at x.

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`\sf \large{x = v_(xi)} * t \\ \\ \sf \large{x = v_i * \cos {\theta}_(i) * t}`

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`\large\textsf{\underline{Now substituting the required values}}`

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`\sf x = 300 * \cos 55{\degree} * 12 \\ \\ \sf x = 100 * 0.5756 * 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}`

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The vertical position of projectile at y.

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`\sf \large y = v_(yi) * t - ((1)/(2) * g * {t}^(2)) \\ \\ \sf \large y = v_i * \cos \theta * t - (1)/(2) g {t}^(2)`

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`\textsf{ \large {\underline{Now substituting the required values}} }`

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`\sf y = 100 * \cos55{ \degree} * 12 - (1)/(2) * 9.80 * {12}^(2) \\ \\ \sf y = 100 * 0.8192 * 12 - 0.5 * 9.8 * 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}`

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Henceforth, the distance at horizon is 688.32m and at vertical is 277.44m.