Thebaseball team practices using an automatic pitching machine depicted. The mac

Question

Thebaseball team practices using an automatic pitching machine depicted. The machine rests atop a tripod of height h = 2.30m and launches balls horizontally toward the waiting batter. Inside the machine, a ball mass m = 0.160kg, begins from rest at the back of a tube, length L = 0.500m. The machine then accelerates the ball down the tube by applying a constant force, causing it to emerge from the front of the tube at velocity v0 = 22.0m/s. For the purposes of this problem, neglect air resistance and treat the ball as a point mass.

a. What is the magnitude of the force F that the bowling machine applies to a ball while accelerating it in the tube?

b. Once it leaves the tube, what distance d (measured from the spot on the ground directly below the front of the tube) does the cball travel in the horizontal direction before landing on the ground?

c. Let vf be the speed of the ball in the instant just before it hits the ground. What is the difference (vf - v0) (i.e. how much faster is the ball traveling just before it hits the ground than when it left the end of the tube)?

Explanation / Answer

A),force (F)=massx(v0^2)/(2*length of barrel)

=(0.16*22^2)/(2*0.5)

=77.44joule

B)distance (d)=v0*time taken to reach ground

=22*sqrt((2*h)/g)

=22*sqrt((2*2.3)/9.8)

=15.07m

C)final speed(vf)=sqrt(v0^2+(gt)^2)

Here t=sqrt((2h/g)=sqrt((2*2.3)/9.8)=0.685seconds

Vf=sqrt(22^2+(9.8*0.685)^2)

=23.001m/s

Thebball hits the ground with 23.001m/s

Vf-v0=23.001-22

=1.001 m/s

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