Question

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 < 130, 0, −3 > ft/s?

Answer 1

Answer:

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

Step-by-step explanation:

Given:

V_0 = < 130 , 0 , -3 > ft / s

a = < 0 , 32 , 8 > ft / s^2

S_plate = < 60 , z , 0 > ft

S_0 = < 0 , 6 , 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