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A ball is thrown at an angle of 40 degrees above the horizontal. The horizontal component of the baseball's initial velocity is 12.0 meters per second. What is the magnitude of the ball's initial velocity?

Respuesta :

jennyurbay1
 Let,
initial velocity = v m/sec,

Angle, x = 40 degrees,

horizontal-componant = v.cos(x) = 12 m/sec,
OR,
v = 12 / cos(40) = 12/0.766 = 15.67 meters/sec >================< ANSWER Source(s): Fazaldin A · 4 years ago
asifejaz20

Answer:

Magnitude of initial velocity = Vi = 15.7 m/s

Explanation:

Horizontal component of velocity = Vi × Cosθ = 12   ……….. (i)

Where,  

θ  = Angle at which ball is thrown above the horizontal = 40 degrees  

Put θ = 40 degrees in equation (i),

Vi × Cos(40)= 12

Vi = 12/ Cos(40)

Vi = 12/0.766

Vi = 15.66 m/s

Vi = 15.7 m/s

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