A 16-ft ladder leans against a wall. The bottom of the ladder is 5ft from the wa

Question

A 16-ft ladder leans against a wall. The bottom of the ladder is 5ft from the wall at time t=0 and slides away from the wall at a rate of 3ft/s. Find the velocity at the top of the ladder at time t=1.

a)The provided answer for this question is -sqrt3 or approximately -1.732 ft/s. Check that the provided answer is correct.

b) The speed of sound at sea-level using the standard atmosphere is about 340.29 meters per second. There are 3.280840 feet in one meter. Using the assumptions of this model, find the angle between the ladder and the ground at the time that the top of the ladder breaks the speed of sound.

c)The speed of light is about 299,792,458 meters per second. There are still 3.280840 feet in one meter. Using the assumptions of this model, find the angle between the ladder and the ground at the time the top of the ladder moves at the speed of light.

Explanation / Answer

L² = x² + y²

L = 16

At t = 0, x(0) = 5

since dx/dt = 3 ft/sec, then x(1) = 8

Solve for y position at t = 1

y(1) = ( 16² - 8² ) = ( 256 - 64 ) = 192

Find the time derivative of L² = x² + y²

0 = 2x dx/dt + 2y dy/dt

Solve for dy/dt

dy/dt = -2x/2y dx/dt = [ - x(t) / y(t) ] dx/dt

Plug in knowns at t = 1

dy/dt = ( -8/192 )( 3 ft/sec )

dy/dt = -24/192 ft/sec = -1.732 ft/sec

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Let the height of the ladder = h
Let the angle between the ladder and the ground =

The distance of the bottom of the ladder from the wall = 3t + 5
= arccos[(3t + 5)/16]

h^2 = 16^2 - (3t + 5)^2
h = sqrt[256 - (3t + 5)^2]
dh/dt = - 6(3t + 5)/{2 sqrt [256 - (3t + 5)^2]}
= - 3(3t + 5)/sqrt[256 - (3t + 5)^2]

340.29 m/s = (340.29 m/s) * (3.280840 ft/m) = approx. 1,116.44 ft/s

Let this value be B

We require 3(3t + 5)/sqrt[256 - (3t + 5)^2] = B
9(3t + 5)^2/[256 - (3t + 5)^2] = B^2
9(3t + 5)^2 = 256B^2 - (3t + 5)^2 B^2
(3t + 5)^2 (B^2 + 9) = 256 B^2
(3t + 5)^2 = 256 B^2/(B^2 + 9)
3t + 5 = 16 B / sqrt (B^2 + 9)
(3t + 5)/16 = B / sqrt (B^2 + 9)

So when the top of the ladder breaks the sound barrier, the angle between the ladder and the ground is given by

= arccos[(3t + 5)/16]
= arccos[B/sqrt(B^2 + 9)]
= approx. arccos 0.999996389
= 0.154° or 0.00269 radians

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