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12345 [234]
3 years ago
12

Completenlo por favor

Physics
1 answer:
Ainat [17]3 years ago
3 0

La velocidad \mathbf v del objeto al tiempo t es descrito por

\mathbf v(t)=v_x(t)\,\mathbf i+v_y(t)\,\mathbf j

donde

\begin{cases}v_x(t)={v_x}_0+a_xt\\v_y(t)={v_y}_0+a_yt\end{cases}

El objeto no tiene aceleración horizontal, pues a_x=0 y v_x está determinado exactamente por su velocidad inicial en la dirección horizontal. En la dirección vertical, la gravedad es la única fuerza que actúa en el objeto, pues a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}. Entonces, la velocidad después de 3\,\mathrm s satisface

{v_x}_0\,\mathbf i+\left({v_y}_0+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3\,\mathrm s)\right)\,\mathbf j=20\,\mathbf i-4\,\mathbf j

Inmediatamente, vemos que {v_x}_0=20\,\dfrac{\mathrm m}{\mathrm s} y podemos encontrar que {v_y}_0=25.43\,\dfrac{\mathrm m}{\mathrm s}.

Su posicíon es descrita al tiempo t por

\mathbr r(t)=r_x\,\mathbf i+r_y\,\mathbf j

con

\begin{cases}r_x={r_0}_x+{v_0}_xt+\dfrac12a_xt^2\\\\r_y={r_0}_y+{v_0}_yt+\dfrac12a_yt^2\end{cases}

Si la posición inicial del objeto es el origen, suponemos que \mathbr r(0)=\mathbf 0, y además tenemos

\mathbf r(t)=\left(20\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\mathbf i+\left(\left(25.43\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right)\,\mathbf j

Queremos determinar el máximo de r_y. Encontramos que {r_y}_{\mathrm{max}}\approx32.9\,\mathrm m cuando t\approx2.59\,\mathrm s.

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