If the splash is heard 1.17 seconds later, what was the initial speed of the rock? take the speed of sound in the air to be 343 m/s.|v0|=nothingm/s.

Incomplete question.The complete question is here

A rock is thrown downward into a well that is 9.00m deep.

If the splash is heard 1.17 seconds later, what was the initial speed of the rock? Take the speed of sound in the air to be 343 m/s.

v₀= 2.263 m/s(initial speed)

Explanation:

Given Data

well is = 9.00m deep

time=1.17 seconds

speed of sound =343 m/s

To find

Speed of rock=?

Solution

as

gravity=9.8 m/s²

Time the splash reaches=d/v = 9/343= 0.026239s  

Rock fall time= 1.17-0.0262= 1.1438s  

thrown downward:

h=h₀-v₀t+1/2gt²eq(i)

Where h=0m

h₀=9m

t=1.1438s

g=-9.8m/s²

put the values in eq(i)

0= 9-(v₀×1.1438)+{1/2(-9.8)×(1.1438)²}

1.1438v₀= 2.589  

v₀= 2.263 m/s(initial speed)

u = 2.264m/s

Therefore the initial speed of the rock is 2.264m/s downwards.

Question:

A rock is thrown downward into a well that is 9.00m deep. If the splash is heard 1.17 seconds later, what was the initial speed of the rock? Take the speed of sound in the air to be 343 m/s.

Explanation:

Given;

Speed of sound in air vs= 343m/s

Time taken to hear the sound t= 1.17s

Distance travelled by rock is d = 9.00m (travelled downward)

The total time taken to hear sound t can be written as;

t = t1 +t2 1

Where

t1 is the time taken for the rock to reach the bottom of the well and

t2 is the time taken for the sound to travel back to the top of the well.

t2 = distance/velocity of sound = d/vs = 9/343 = 0.0262s

From equation 1

t1 = t - t2

t1 = 1.17 - 0.0262

t1 = 1.1438

For the rock traveling, down the well the formula for distance travelled by the rock is;

d = ut + 0.5gt^2

u = (d -0.5gt^2)/t

For this case t = t1

g = 9.8m/s2 acceleration due to gravity

d = 9

u is the initial speed of rock.

u = (9 - 0.5×9.8×1.1438^2)/1.1438

u = 2.5894/1.1438

u = 2.264m/s

Therefore the initial speed of the rock is 2.264m/s downwards.