Question

A softball is thrown from the origin of an x-y coordinate system with an initial speed of 14 m/s at an angle of 27∘ above the horizontal. a) What is the initial y component of the velocity? b) What is the y position of the ball after 0.5 seconds?

Answer 1

A). To find initial y component of velocity we will split the initial velocity, u(=14m/s) into horizontal and vertical component as $\fn_cm \large ucos\Theta$ and $\fn_cm \large usin\Theta$ respectively. y component $\fn_cm \large = usin\Theta$ where $\fn_cm \large \Theta = 27$ ∘

Therefore u_y = 14 × sin(27) = 14 × 0.5439 = 6.3558 m/s.  $\fn_cm \large \therefore$ u_y = 6.3558 m/s

B). To find the y position of ball after 0.5 seconds, we simply multiple the y complemento of velocity with given time.

Therefore, y position of ball after 0.5 seconds = 6.3558 × 0.5 = 3.178 m