The formula v=fλ can be used here.
326=2500*λ
Note the 2500 as 2.5kHz is 2.5 thousand Hz.
λ = 326/2500
= 0.1304m = 0.130m
When air is blown into the open pipe,
L = 
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ=
=3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f= 
⇒f=
≈115 Hz
Answer:
12.7 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 56.7 Km/hr
Maximum height (h) =..?
First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:
Initial velocity (m/s) = 56.7 x 1000/3600
Initial velocity (m/s) = 15.75 m/s
Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:
Initial velocity (u) = 15.75 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
v = u – gt (since the ball is going against gravity)
0 = 15.75 – 9.8 × t
Rearrange
9.8 × t = 15.75
Divide both side by 9.8
t = 15.75/9.8
t = 1.61 secs.
Finally, we shall determine the maximum height as follow
h = ½gt²
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 1.61 secs.
Height (h) =..?
h = ½gt²
h = ½ × 9.8 × 1.61²
h = 4.9 x 1.61²
h = 12.7 m
Therefore, the maximum height reached by the ball is 12.7 m
Use the equation in the physics reference sheet. You can use the distance formula