An arrow is shot into the air at an elevation angle of 25 degrees. If its projection velocity is 55 m/s. What is the vertical component of its initial velocity? What is its horizontal component?

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Answer 1 · 2015-07-04T15:58:30

Vertical component: 23 m/s
Horizontal component: 50. m/s

For a given launch angle #theta#, the vertical and horizontal components of the initial velocity can be written as

#{(v_(oy) = v_0 * sin(theta)), (v_(0x) = v_0 * cos(theta)) :}#

In your case, the two components of the initial speed will be

#v_(oy) = "55 m/s" * sin(25^@) = color(green)("23 m/s")#

and

#v_(0x) = "55 m/s" * cos(25^@) = color(green)("50. m/s")#

Both values are rounded to two sig figs, the number of sig figs you gave for the arrow's initial velocity.