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

Answer

688.32m and 277.44m

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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$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

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

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

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The vertical position of projectile at y.

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$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

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

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$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$

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