Question

(a)A soccer player kicks a rock horizontally off a 35 m high cliff into a pool of water. If the player hears the sound of the splash 2.83 s later, what was the initial speed given to the rock (in m/s)? Assume the speed of sound in air is 343 m/s.(b)What If? If the temperature near the cliff suddenly falls to 0°C, reducing the speed of sound to 331 m/s, what would the initial speed of the rock have to be (in m/s) for the soccer player to hear the sound of the splash 2.83 s after kicking the rock?

Answer 1

Answer:

a)v=15.83 m/s

b)v=14.89 m/s

Explanation:

a)

In y direction

$h=v_yt+\dfrac{1}{2}gt^2$

Given that h=35 m and initial velocity is zero.

So

$35=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=2.67 sec

It means that rock hit water after 2.61 sec.

Given that player hears after 2.83 sec.

Δt=2.83 - 2.67 s

Δt=0.16 sec

So distance travel by sound wave

d= C x Δt

C= 343 m/s

d= 343 x 0.16

d=54.88 m

From diagram

$d^2=x^2+h^2$

$x=\sqrt{d^2-h^2$

$x=\sqrt{54.88^2-35^2$

x=42.27 m

Now

x = v x  t

v=42.27/2.67

v=15.83 m/s

b)

Now sound speed have been changed

So new d

d=331 x 0.16

d=52.96 m

New x

$x=\sqrt{d^2-h^2$

$x=\sqrt{52.96^2-35^2$

x=39.75

So new velocity v

v=x/t

v=39.75/2.67

v=14.89 m/s