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: f_rocket, x = C/V_x where Cis a constant equaling 200 kg m^-2 s^_3. The total distance the ball travels before it hits the catcher's mitt is 60 ft and 6 inches. Gravity is 9.807 m s^-2. Compute: The final velocity in the x-direction in [mile/hr] 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

weight of baseball= 5.1 oz =144.583 g=0.144 kg

F=C/Vx = 200 /Vx

so, mdv/dt = 200/Vx

so, dv/dt = 200/mVx

so, vdv/ds =200/mvx

so, v^2dv=(200/m) ds

so, v^3-vo^3= (600/m) s

vo= 90 niles/hr= 40.2336 m/s s=60.5 feet =18.44 m

so, v^3= (40.2336)^3 + (600/m) s = 65127.8 + 4166.66=69294.46

so, v= 40.92 m/s= 91.537 miles/hr

b. dv/dt = 200/mVx

so, vdv=(200/0.144) dt

so, V^2-vo^2= (400/0.144) t

so, 40.92^2 - 40.2336^2 = (400/0.144)t

so, t = 20 ms

c. vz= 9.807 x 20 x 10^-3= 0.1966 m/s =0.4397 miles/hr

d. sz= 1/2 x 9.807 x (0.02)^2= 0.00196 m =0.0064 ft

e.initial ke= 1/2 x 0.144 x 40.2336^2= 116.626 j

final ke= 1/2 x 0.144 x 40.92^2= 120.56 j

loss in final pe= -0.144 x9.807 x 0.00196= -0.00276 j

f. work done = change in energy=(120.56-0.00276 -116.626)=3.93124 j

g. air resistance and its effect is negligible

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