Question

An archer fires an arrow with a velocity of 200 m/s at an angle of 55 degrees above horizontal?a. What is the horizontal component of its initial velocity?b. What is the vertical component of its initial velocity?C.What is the maximum height attained by the arrow?d. How long does it take the arrow to reach that height?e.What is the total amount of time that it's in the air?f. How far away does it strike the ground?g. What is the horizontal component of its velocity just prior to impact?h. What is the vertical component of its velocity just prior to impact?i. What is the magnitude of its velocity just prior to impact?j. What is the direction of its velocity just prior to impact?

Answer 1

Given data:

The initial velocity of an arrow is

`v=200\text{ m/s}`

The angle above the horizontal is

`\theta=55\degree`

Part (a)

The horizontal component of its initial velocity can be given as,

`v_H=v\cos \theta`

Substituting the values in the above equation, we get:

`\begin{gathered} v_H=200\text{ m/s}*\cos 55\degree \\ =114.715\text{ m/s} \end{gathered}`

Thus, the horizontal component of its initial velocity is 114.715 m/s.