Question

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?

Answer 1

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

Answer 2

Magnitude of initial velocity = Vi = 15.7 m/s

Step-by-step 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