QUESTION IMAGE
Question
an arrow is shot from a bow at an angle of 35° above the horizontal with an initial speed of 50. m/s. what are the arrows horizontal (x) and vertical (y) components? (use trigonometry to answer, but remember to draw a diagram to help your understanding.)
x = m/s
y = 50.
29
41 m/s
Explicación:
Paso 1: Encontrar la componente horizontal
La componente horizontal de la velocidad inicial $v_{0x}$ se calcula usando la fórmula $v_{0x}=v_0\cos\theta$, donde $v_0 = 50\ m/s$ y $\theta = 35^{\circ}$.
$v_{0x}=50\cos(35^{\circ})$
$v_{0x}\approx50\times0.819 = 40.95\approx41\ m/s$
Paso 2: Encontrar la componente vertical
La componente vertical de la velocidad inicial $v_{0y}$ se calcula usando la fórmula $v_{0y}=v_0\sin\theta$, donde $v_0 = 50\ m/s$ y $\theta = 35^{\circ}$.
$v_{0y}=50\sin(35^{\circ})$
$v_{0y}\approx50\times0.574 = 28.7\approx29\ m/s$
Respuesta:
$x = 41\ m/s$
$y = 29\ m/s$
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Explicación:
Paso 1: Encontrar la componente horizontal
La componente horizontal de la velocidad inicial $v_{0x}$ se calcula usando la fórmula $v_{0x}=v_0\cos\theta$, donde $v_0 = 50\ m/s$ y $\theta = 35^{\circ}$.
$v_{0x}=50\cos(35^{\circ})$
$v_{0x}\approx50\times0.819 = 40.95\approx41\ m/s$
Paso 2: Encontrar la componente vertical
La componente vertical de la velocidad inicial $v_{0y}$ se calcula usando la fórmula $v_{0y}=v_0\sin\theta$, donde $v_0 = 50\ m/s$ y $\theta = 35^{\circ}$.
$v_{0y}=50\sin(35^{\circ})$
$v_{0y}\approx50\times0.574 = 28.7\approx29\ m/s$
Respuesta:
$x = 41\ m/s$
$y = 29\ m/s$