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Assuming that the acceleration due to gravity is 32.2 ft/s2 and air resistance is negligible, a projectile is launched at 56° above the horizontal with an initial velocity of 36 ft/s. The launch and landing sites are at the same elevation.

What is the projectile’s range?

Respuesta :

MathPhys

Answer:

37 ft

Explanation:

First find the air time:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 0 + (36 sin 56°) t + ½ (-32.2) t²

0 = 29.8 t - 16.1 t²

0 = t (28.8 - 16.1 t)

t = 1.85 s

Now find the range:

x = x₀ + v₀ t + ½ at²

x = 0 + (36 cos 56°) (1.85) + ½ (0) (1.85)²

x = 37.3 ft

Rounded to two sig-figs, the range is 37 feet.

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