contestada

A projectile is launched at an angle of 60° from the horizontal and at a velocity of
12.0 m/s. What is the horizontal velocity of the projectile? *

Respuesta :

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

The horizontal velocity of the projectile is 6.0m/s

If a projectile is launched at an angle from the horizontal and at a velocity v, the horizontal velocity of the projectile is expressed as:

[tex]v_x =vcos \theta[/tex]

Given the following parameters

v = 12.0m/s

[tex]\theta=60^0[/tex]

Substitute the given parameters into the formula to have:

[tex]v_x=12.0cos60\\v_x=12.0(0.5)\\v_x=6.0m/s\\[/tex]

Hence the horizontal velocity of the projectile is 6.0m/s

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