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3 years ago

11

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?

2 answers:

motikmotik3 years ago

8 0

<span> 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 </span> Source(s): Fazaldin A <span> · 4 years ago </span>

Andru [333]3 years ago

7 0

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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Answer:

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