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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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Nash

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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. Calculamos

v_0^2

25.5^2 = 650.25

2. Dividimos el resultado por

R = \frac{650.25}{9.8} = 66.35 \, m

Respuesta:

R = 66.35 \, m

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The maximum range (on flat ground) of a projectile launched at 25.5 m/s is

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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