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A projectile is fired horizontally from a gun that is 50.0 m above flat ground, emerging from the gun with a speed of 210 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Respuesta :

gilcarusolautaro

Answer:

(a): The projectile remain in the air t=3.19 seconds.

(b): The projectile strikes the ground at d= 669.9 meters in horizontal distance from the firing point.

(c): The magnitude of the vertical component of its velocity is Vy= 31.262 m/s when the projectile strikes the ground.

Explanation:

g= 9.8 m/s²

h1= 50 m

Vx= 210 m/s

h= g*t²/2

[tex]t=\sqrt{\frac{h1*2}{g} }[/tex]

t= 3.19 s  (a)

d= Vx * t

d= 669.9 m (b)

Vy= g*t

Vy= 31.262 m/s  (c)

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