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A 42.0 kg projectile is launched horizontally from the top of a cliff, 50 meters above the ground. The projectile’s initial speed is 30 m/s. What is the velocity right before it hits the ground

Respuesta :

Answer:

Velocity of the projectile just before it will hit the surface is 43.4 m/s

Explanation:

In x direction there is no acceleration

so horizontal velocity of the projectile is always constant

So we have

[tex]v_x = 30 m/s[/tex]

now in Y direction its motion is free fall motion under gravity

so we have

[tex]v_y^2 = 0^2 + 2(9.81)(50)[/tex]

[tex]v_y = 31.3 m/s[/tex]

so we have

[tex]v = \sqrt{v_y^2 + v_x^2}[/tex]

[tex]v = \sqrt{30^2 + 31.3^2}[/tex]

[tex]v = 43.4 m/s[/tex]

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