Two window washers start at the heights shown. (A: 21 ft high rising 8 in per second. The other is 50 feet high descending 11inc

Question

2 months ago

9

hes per second) one is rising one is descending. How long does it take for the two window washers to reach the same height? Explain

2 answers:

Leona [12.6K] · Answer 1 · 2026-02-17T02:36:29+03:00

To address this problem, let's start by formulating the general motion equation along the vertical direction.

This gives us:

$h = \frac {1} {2} gt ^ 2 + vo * t + h0$

Where,

For the first individual:

$h1 = \frac {1} {2} gt ^ 2 + \frac {8} {12} * t + 21$

For the second individual:

$h2 = \frac {1} {2} gt ^ 2 - \frac {11} {12} * t + 50$

When both individuals reach the identical altitude, the following holds:

$h1 = h2\\$

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

Rearranging results in:

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

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

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

Solving for time:

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

Result:

The two window washers reach the same height after 18.31 seconds.

Svet_ta [12.7K] · Answer 2 · 2026-02-17T02:37:15+03:00

Result:

After 18.32 seconds, the two window washers will be at the same level.

Detailed steps:

Assume h denotes the height where window washer A meets window washer B.

Washer A moves upward at 8 inches per second.

Initially, A is positioned at 21 feet, which equals 252 inches (21 × 12).

If they meet after t seconds, then the height of A at that instant is:

h = 252 + 8t ---------(1)

Washer B starts at 50 feet, or 600 inches (50 × 12).

B moves downward at 11 inches per second.

Therefore, at time t, B's height is:

h = 600 - 11t --------(2)

Setting these equal gives:

252 + 8t = 600 - 11t

8t + 11t = 600 - 252

19t = 348

t = $\frac{348}{19}$

t = 18.32 seconds