Question

A 6m ladder leans against a wall. The bottom of the ladder is 1.3m from the wall at time =0sec and slides away from the wall at a rate of 0.3m/s.

Find the velocity of the top of the ladder at time =2 (take the direction upwards as positive).


(Use decimal notation. Give your answer to three decimal places.)

Answer 1

- 0.100

Explanation:

Length of the ladder, H = 6 m

Distance at the bottom from the wall, B = 1.3 m

Let the distance of top of the ladder from the bottom at the wall is P

Thus,

from Pythagoras theorem,

B² + P² = H² .

or

B² + P² = 6² ..............(1) [Since length of the ladder remains constant]

at B = 1.3 m

1.3² + P² = 6²

or

P² = 36 - 1.69

or

P² = 34.31

or

P = 5.857

Now,

differentiating (1)

`2B((dB)/(dt))+2P((dP)/(dt))=0`

at t = 2 seconds

change in B = 0.3 × 2= 0.6 ft

Thus,

at 2 seconds

B = 1.3 + 0.6 = 1.9 m

therefore,

1.9² + P² = 6²

or

P = 5.69 m

on substituting the given values,

2(1.9)(0.3) + 2(5.69) ×
`((dP)/(dt))` = 0

or

1.14 + 11.38 ×
`((dP)/(dt))` = 0

or

11.38 ×
`((dP)/(dt))` = - 1.14

or

`((dP)/(dt))` = - 0.100

here, negative sign means that the velocity is in downward direction as upward is positive