Question

Two window washers start at the heights shown. (A: 21 ft high rising 8 in per second. The other is 50 feet high descending 11inches per second) one is rising one is descending. How long does it take for the two window washers to reach the same height? Explain

Answer 1

For this case, the we have that the generic equation of motion is:

We have then:

`y = \frac {1} {2} gt ^ 2 + vo * t + yo`

Where,

• g: acceleration of gravity
• vo: initial speed
• yo: initial height

First equation:

`y1 = \frac {1} {2} gt ^ 2 + \frac {8} {12} * t + 21`

Second equation:

`y2 = \frac {1} {2} gt ^ 2 - \frac {11} {12} * t + 50`

For same height we have:

`y1 = y2\\`

`\frac {1} {2} gt ^ 2 + \frac {8} {12} * t + 21 = \frac {1} {2} gt ^ 2 - \frac {11} {12} * t + 50`

Clearing time:

`\frac {8} {12} * t + 21 = - \frac {11} {12} * t + 50`

`\frac {8} {12} * t + \frac {11} {12} * t = 50 - 21`

`\frac {19} {12} * t = 29`

`t = 29 (\frac {12} {19})\\t = 18.31s`

Answer:

it takes 18.31s for the two window washers to reach the same height

Answer 2

18.32 seconds

Explanation:

Given: one washer is 21 ft high rising 8 in per second. The other is 50 feet high descending 11inches per second.

To Find: How long does it take for the two window washers to reach the same height

Solution:

Initial height of first washer
`=21`
`\text{feets}`

`1\text{feet}=12\text{inch}`

height of first tower
`=21*12=252\text{inches}`

ascending rate of first tower
`=8`
`\text{in}/ \text{s}`

equation for height of first washer after
`\text{x}` seconds

`=252+8\text{x}`

Initial height of second washer
`=50`
`\text{feets}`

`1\text{feet}=12\text{inch}`

height of second tower
`=50*12=600\text{inches}`

descending rate of second tower
`=11`
`\text{in}/ \text{s}`

equation for height of second washer after
`\text{x}` seconds

`=600-11\text{x}`

when height of both washers are equal

`252+8\text{x}=600-11\text{x}`

`19\text{x}=348`

`\text{x}=(348)/(19)`

`\text{x}=18.32\text{s}`

It will take 18.32s to both washers to reach the same height