A simple model to describe the curve of a baseball assumes the spin of the ball produces a constant sideways acceleration (in the y-direction) of c ft/s2. Suppose a pitcher throws a curve ball with c = 8 ft/s2. How far does the ball move in the y-direction by the time it reaches home plate, assuming an initial velocity of ft/s?

S_y, plate = - 0.5325 ft (further away from batter)

Step-by-step explanation:

Given:

V_0 = ft / s

a = ft / s^2

S_plate = ft

S_0 = ft

The time taken t by the ball to reach home plate:

S_x, plate = S_o, x + V_o, x * t

60 = 0 + 130*t

t = 60 / 130

t = 0.4615 s

The distance traveled by the ball in y-direction when it reaches home plate at t = 6 / 13 s:

S_y, plate = S_o, y + V_o, y * t + 0.5*a_y*t^2

S_y, plate = 0 - 3 * (0.4615) + 0.5 * 8 * (0.4615) ^2

S_y, plate = - 0.5325 ft