A stunned Batman finds himself in a vault on a high ledge. In an attempt to trap

Question

A stunned Batman finds himself in a vault on a high ledge. In an attempt to trap him, the Joker runs out of the vault while the door closes. Batman fires a grappling hook into the ceiling at an angle, , attached to a line of length, L, and manages to pull himself off the ledge swinging down to the floor below. At the very bottom of the swing, he lets go of the line to the grappling hook in an attempt to slide the rest of the way out of the door. Find an equation for how far Batman slides, x, as a function of L, and k, the coefficient of kinetic friction between Batman and the floor. If the door is 6.00 meters away from the bottom of the swing and the length of the line is 3.00 meters, the angle 70 degrees and the coefficient of friction 0.250 did Batman make it out? If he did, how fast was he sliding when he went through the door?

Explanation / Answer

here,

let the speed of Batman at the bottom be v

using conservation of energy

0.5 * m * v^2 = m * g * L * ( 1 - cos(theta) )

v = sqrt( 2 * g * L * ( 1 - cos(theta))

the accelration due to friction , a = - uk * g

v^2 - u^2 = 2 * a*x

x = ( 2 * g * L * ( 1 - cos(theta)) /( 2 * uk *g)

x = ( 2 * L * ( 1 - cos(theta)) /( 2 * uk )

the expression for x is x = ( 2 * L * ( 1 - cos(theta)) /( 2 * uk )

L= 3 m , theta = 70 degree , uk = 0.25

then x = ( 2 * 3 * ( 1 - cos(70)) /( 2 * 0.25 )

x = 7.9 m

the door is 6 m from the bottom , so ,

he made it out

let his speed at the door be u

using third equation of motion

u^2 - ( 2 * g * L * ( 1 - cos(theta)) = - 2 * ( uk * g) * ( 6)

u^2 - ( 2 * 9.8 * 3 * ( 1 - cos(70)) = - 2 * ( 0.25 * 9.8) * ( 6)

u = 3.05 m/s

the speed of Batman when he went through the door is 3.05 m/s

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