Respuesta :
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.
Answer:
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+1/2gt²...................eq(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)
Answer: 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.
