A baseball is thrown with a horizontal velocity of 40.20 m/s and no vertical velocity. If the ball leaves the pitcher's hand at a height of 2.000 m above the ground, how far will the ball travel before it hits the ground? (Air resistance is negligible.)

A. 12.68 m

B. 16.48 m

C. 25.73 m

D. 51.46 m

Option (c) is correct.

Explanation:

initial velocity along vertical=0

velocity along horizontal Vx=40.20 m/s

acceleration along the vertical=9.8 m/s²

displacement along vertical, h=2 m

using the kinematic equation

h = vi*t + 1/2 g t²

2=0 (t) + 1/2 (9.8) t²

t²=0.408

t=0.639 s

now the displacement along vertical = x = Vx * t

x=40.20 (0.639)

x=25.73 m