always use significant figure rules. remember that these rules apply to all numbers that are measurements…

always use significant figure rules. remember that these rules apply to all numbers that are measurements. time links motion in both the x and y direction. an arrow is shot from a bow at 20.0 m/s at an angle of 65° above the horizontal. the arrow leaves the bow at a height of 1.80 m. at what height will it strike a wall that is 10.0 m away? (remember to connect the two motions of the arrow using time.) y = m 16.4 1.80 21.5

always use significant figure rules. remember that these rules apply to all numbers that are measurements. time links motion in both the x and y direction. an arrow is shot from a bow at 20.0 m/s at an angle of 65° above the horizontal. the arrow leaves the bow at a height of 1.80 m. at what height will it strike a wall that is 10.0 m away? (remember to connect the two motions of the arrow using time.) y = m 16.4 1.80 21.5

Answer

Explicación:

Paso 1: Encontrar la componente horizontal de la velocidad inicial

La velocidad inicial $v_0 = 20.0$ m/s y el ángulo $\theta=65^{\circ}$. La componente horizontal de la velocidad inicial es $v_{0x}=v_0\cos\theta$. $v_{0x}=20.0\cos(65^{\circ})\approx20.0\times0.423 = 8.47$ m/s

Paso 2: Calcular el tiempo que tarda en llegar a la pared

La distancia horizontal $x = 10.0$ m. Usando la ecuación $x = v_{0x}t$, despejamos $t$. $t=\frac{x}{v_{0x}}=\frac{10.0}{8.47}\approx1.18$ s

Paso 3: Encontrar la componente vertical de la velocidad inicial

La componente vertical de la velocidad inicial es $v_{0y}=v_0\sin\theta$. $v_{0y}=20.0\sin(65^{\circ})\approx20.0\times0.906 = 18.12$ m/s

Paso 4: Calcular la altura en la que golpea la pared

Usamos la ecuación de movimiento vertical $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, donde $y_0 = 1.80$ m y $g = 9.8$ m/s². $y=y_0+v_{0y}t-\frac{1}{2}gt^{2}$ $y = 1.80+18.12\times1.18-\frac{1}{2}\times9.8\times(1.18)^{2}$ $y = 1.80 + 21.38 - 6.82$ $y=16.4$ m

Respuesta:

$16.4$ m