Question

A stone is dropped from a 50 m tall building into a pond. When is sound of splash heard at the top? (g=10m/s2 , speed of sound in air= 340m/s)?​

Answer 1

A stone is dropped from a 50 m tall building into a pond.

Initially stone is at rest. So, the initial velocity of the stone is 0 m/s. As given in question that, stone is dropped from a 50 m tall building. Means, the height (h or s) is 50 m.

We have to find the sound of splash heard at the top. (In short we have to find time i.e. t).

USING SECOND EQUATION OF MOTION,

s = ut + 1/2 at²

Substitute the known values

50 = 0(t) + 1/2 at²

Here, a is acceleration and its value is 10 m/s².

50 = 0 + 1/2 × 10 × t²

50 = 5t²

10 = t²

√10 = t

t = 3.16 sec

Also,

Distance = Speed × time

And given speed of sound in air is 340 m/s and distance is 50 m (from which the stone is dropped).

50 = 340 × time

50/340 = time

0.14 = time

time = 0.14 sec

Total time taken by sound = (3.16 + 0.14) sec = 3.30 sec

Answer 2

Given :-

• distance travelled by stone, s = 50 m
• acceleration due to gravity, g = 10 m/s²
• velocity of sound in air = 340 m/s

To find :-

time after which sound of  splash will be heard at the top = T

Solution :-

initial velocity of stone ,u = 0  m/s

Let,

time taken by stone to reach the base of building = t₁

so,

Using second eqn of motion :

$\boxed{\bf{s=ut+ \frac{1}{2}at^2}}$

where ,

s is the distance travelled , u is the initial velocity , t is the time taken to travel distance s , a is the acceleration .

putting values

$\tt{50=0(t_1)+\frac{1}{2}(10)(t_1)^2}$

$\tt{50 = \frac{10(t_1)^2}{2}}$

$\tt{100=10(t_1)^2}$

$\boxed{\tt{t_1=\sqrt{10}\:s=3.162\:s}}$

now,

Sound will have to travel distance  = 50 m

speed of sound in air = 340  m/s

Hence,

time taken by sound to reach at top ,

   t₂ = 50 / 340  

  $\boxed{\tt{t_2=0.147\:s}}$

So,

time after which sound of splash will be heard at top ,

T = t₁ + t₂

T = 3.162 s + 0.147 s

$\boxed{\boxed{\bf{T=3.309\:s}}}$

Hence,

sound of splash will be heard at top after 3.309 sec.