A pitcher releases a baseball weighing 5.1 oz approximately 7 feet above the gro

Question

A pitcher releases a baseball weighing 5.1 oz approximately 7 feet above the ground at 90 miles per hour (x-direction initial velocity; no z-direction initial velocity). Instantaneously after the ball leaves her hand, an internal rocket in the ball fires providing the following force further propelling the ball forward: the catcher's mitt is 60 ft and 6 inches. Gravity is 9.807 m s'2. Compute: The total time the ball travels in the air in [ms] The final velocity in the z-direction in [mile/hr]. The distance the ball drops in the z-direction in [ft] The initial and final KE and PE in [J]. What do you notice about the Conservation of Energy in Mechanics? The work done by the rocket on the ball in [J]. What is assumed throughout this analysis? Show all calculations and unit conversions otherwise no points will be given.

Explanation / Answer

Mass of Ball = 5.1 oz = 0.145 Kg
(a)
Initial Velocity in x direction = 90 mile/hr = 40.23 m/s

Force in x direction = C/Vx = 200/ 40.23 = 4.97 N
m*a = 4.97 N
a = 4.97/0.145
a = 34.3 m/s^2

Now, Distance travelled by ball in x direction = 60 ft & 6 in = 18.288 + 0.1524 = 18.44 m

Final Velocity in x direction,
V^2 = u^2 + 2*a*s
V^2 = 40.23^2 + 2*34.3*18.44
V = 53.69 m/s
V = 120 miles/hr
Final Speed in x direction, Vx = 120 miles/hr

(b)
Total time travelled by ball in air,
V = u + a*t
t = (V-u)/a
t = (53.69 - 40.23)/34.3
t = 0.392 s
t = 392 ms
Total time ball travels in air, t = 392 ms

(c)
Initial Velocity in Z direction = 0
Acceleration = 9.8 m/s^2
Time = 0.392 s

Vz = u + a*t
Vz = 0 + 9.8 * 0.392
Vz = 3.84 m/s
Vz = 8.6 m/h
Final Velocity of ball in Z direction, Vz = 8.6 m/h

(d)
Distance ball drop in Z Direction,
S = u*t + 1/2 * at^2
S = 0 + 1/2 * 9.8 * 0.392^2
S = 0.753 m
S = 2.47 ft

Distance ball drops in Z Direction, S = 2.47 ft

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