In fresh water sound waves travel at 1497m/s at 25 degrees, I'll assume that's the characteristics of the water.
If it's 0.01s then you need to divide the speed by 100 to get the, 14.97, however it gets there and back in that time so you need to halve it.
<u>7.485m</u>
That's the cool thing about free fall. The amount of time it takes to fall remains the same.
In this case, a ball that is simply dropped from rest will fall at the same rate as a ball that had some umph in the horizontal direction.
Answer:
Whats the question/word problem or where is the graph (if included) representing this problem?
Answer:
6.0 m below the top of the cliff
Explanation:
We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

where
u = 0 (it starts from rest)
g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)
h = 24 m is the distance covered
Solving for h,

So the ball thrown upward is launched with this initial velocity:
u = 21.7 m/s
From now on, we take instead upward as positive direction.
The vertical position of the ball dropped from the cliff at time t is

While the vertical position of the ball thrown upward is

The two balls meet when

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

So the distance below the top of the cliff is

Answer:
1/3 times.
Explanation:
Let V₀ be the peak voltage .
IR ( rms ) = ( V₀ / √2 R )
R is value of resistance
IC = ( V₀ ωC / √2 )
( 1 / ωC is reactance of capacitance in ac circuit )

= 
When frequency is tripled angular frequency will also be tripled
hence the ratio
becomes 1/3 times.