Question
The maximum range (on flat ground) of a projectile launched at 25.5 m/s is
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Answer to a math question The maximum range (on flat ground) of a projectile launched at 25.5 m/s is
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Para encontrar el alcance máximo (en terreno llano) de un proyectil lanzado a 25.5 m/s, utilizamos la siguiente fórmula:
R = \frac{v_0^2 \sin (2\theta)}{g}
donde,
v_0 = 25.5 \, m/s
g = 9.8 \, m/s^2
Para el alcance máximo en terreno llano, el ángulo de lanzamiento $\theta$ debe ser 45 grados, lo que maximiza $\sin (2\theta) = \sin (90^\circ) = 1$.
Entonces,
R = \frac{25.5^2 \cdot 1}{9.8}
[SOLUTION]
R = \frac{25.5^2}{9.8}
[STEP-BY-STEP]
1. Calculamosv_0^2
25.5^2 = 650.25
2. Dividimos el resultado porg
R = \frac{650.25}{9.8} = 66.35 \, m
Respuesta:
R = 66.35 \, m
donde,
Para el alcance máximo en terreno llano, el ángulo de lanzamiento $\theta$ debe ser 45 grados, lo que maximiza $\sin (2\theta) = \sin (90^\circ) = 1$.
Entonces,
[SOLUTION]
[STEP-BY-STEP]
1. Calculamos
2. Dividimos el resultado por
Respuesta:
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