Pedro and joanna decide to try a problem. a baseball is thrown from the top of a tall building with an initial velocity of 12.5 m/s from a height of h = 11.5 m above the ground. 1. find its speed when it reaches the ground if its launch angle is −37°. vf = m/s 2. find its speed if it is launched horizontally. vf = m/s

Answer

given,

initial velocity of the base ball = 12.5 m/s

height  of the building = 11.5 m

angle of launch =  −37°

speed when it reaches ground = ?

using kinematics equation

velocity in y direction

v_y = 16.79 m/s

velocity in x- direction

v_x = v cos θ

v_x = 12.5 x cos 37°

v_x = 9.98 m/s

v = 19.53 m/s

b)  if it is launched vertically

initial vertical velocity  = 0

        v_y = 15 m/s

        v_x = 12.5 m/s

v = 19.52 m/s

the answer is to check all of them! they all apply well at least i think so!