2 Batman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor Grabbing

Question

2 Batman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor Grabbing a rope attached to a chandelier, he swings down to grapple with the Joker (mass 70.0 kg) who is standing directly under the chandelier (Assume Batman's center 01 mass moves down 5.0 m. He releases the rope just as he reaches the villain.) a) If Batman pushes himself away from the windowsill with an initial velocity or 5.0 m/s. what is Batman's velocity just before he strikes the Joker's b) Upon impact with the Joker, Batman releases the rope and grabs the villain. What IS their velocity just after impact? Five Point Bonus Question: Although Batman is cool, he is not the best possible superhero. Who is the best superhero? (There is a correct answer to this question.)

Explanation / Answer

a. v1= 5 m/s

Using conservation of momentum

m1v1=m2v2

v2=m1v1/m2=80*5/70=5.71 m/s

b.Gravitational potential energy

U=mgh=8*9.8*5=3920 J

Kinetic energy

T=3920 J =1/2 mv^2

==>v= sqrt(2T/m)=sqrt(2*3920/80)=7.23 m/s

Using conservation of momentum

m1v=m3v3

here, m3 is the mass of Batman and Joker combined

v3=m1v/m3=80*7.23/(80+70)=3.856 m/s

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