Aman drops a rock into a well. (a) the man hears the sound of the splash 2.40 s after he releases the rock from rest. the speed of sound in air (at the ambient temperature) is 336 m/s. how far below the top of the well is the surface of the water? (b) what if? if the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?

Part (a)
Let h =  depth of water surface from the top of the well, m

Because the stone is dropped, its initial velocity is zero.
Let t₁ =  time for the stone to drop, s
Then
h = (1/2)*(9.8 m/s²)*(t₁ s)²
t₁ = √(h/4.9) s = 0.4518√h s

The velocity of sound is 336 m/s.
The time for the sound wave to travel from the water surface to the top of the well is
t₂ = h/336 m/s 

Because the time before the sound is heard is 2.4 s, therefore
t₁ + t₂ = 2.4
0.45181√h + h/336 = 2.4
Multiply through by 336.
151.8082√h + h = 806.4
h - 806.4 = -151.8082√h
h² - 1612.8h + 6.5028 x 10⁵ = 2.3046 x 10⁴h
h² - 2.4659 x 10⁴h + 6.5028 x 10⁵ = 0

Solve with the quadratic formula.
h = 0.5[2.4659 x 10⁴ +/- 2.4606 x 10⁴]
   = 24632 m or 26.5 m

Test the two answers.
When h = 24632,
   t₁ = 0.4518√(24632) = 70.9 s (Not acceptable)
When h = 26.5 m,
   t₁ = 0.4518√(26.5) = 2.3258 s, t₂ = 26.5/336 = 0.0789 s,
   t₁+t₂ = 2.4 s (CORRECT)

 h = 26.5 m

Part (b)
If travel time for the sound is ignored, then
0.4518√h = 2.4
h = (2.4/0.45118)² = 28.2 m
The percent error is 100*[(28.2 - 26.5)/26.5] = 6.42%

The error is  6.4%

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

It is the areas with lines. this may sound a little confusing, but i hope it !