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A projectile is fired into the air at a nonzero angle with the horizontal. When the projectile reaches its maximum height, the speed of the projectile is 50.8% of its original speed. What angle was the projectile originally fired at?

Respuesta :

Answer:

59.5 deg

Explanation:

[tex]v[/tex] = original speed at which the projectile is launched

θ = angle of launch of projectile

[tex]v_{x}[/tex] = component of speed along the horizontal direction =  [tex]v [/tex] Cosθ

At the highest position, the vertical component of velocity becomes zero and there is only horizontal component of velocity, hence

[tex]v_{highest}[/tex] = velocity at the highest point = [tex]v_{x}[/tex] = [tex]v [/tex] Cosθ

it is given that

[tex]v_{highest}[/tex] = 0.508 [tex]v[/tex]

so

[tex]v[/tex] Cosθ = 0.508 [tex]v[/tex]

Cosθ = 0.508

θ = 59.5 deg

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