Kinetic energy of a crane carrying a mass - swing of mass

We will use the conservation of energy to solve this...

The crane is moving with a uniform velocity of 10ft/s. Thus the mass tied to it is also moving with that speed...right ?

what's the Kinetic Energy of this mass ?
Ke = 1/2 m v^2

when the crane suddenly stops, the mass moves forward...and goes up just like a simple pendulum.

And we know, it will go up to a point where its KE is depleted fully. Actually I can't say that as the enrgy is always conserved.

Thus this energy gets converted fully to the potential energy at the top point of the pendulum. See the attached figure.

Now we will write, at the maximum height, KE = PE

ie, 0.5 m v^2 = m g h ....(1)

where h is the max height to which it will rise...

From, (1), 0.5 v^2 = g h (where g = 32ft/sec^2)

Or h = 0.5 v^2/32 = 0.5* 100 /32 = 1.56 ft

We got h, now its time to use the pythagorous theorem to determine d

d = sqrt{30^2 - (30-1.56)^2}

= sqrt(900-808.69) = 9.5 ft

thus the mass will move a horizontal distance of 9.5 ft

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