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Question: A stone dropped from the top of a tower \[525\] m high, falls into a pond at its base on the ground....

A stone dropped from the top of a tower 525525 m high, falls into a pond at its base on the ground. When will the sound of the splash be heard? (speed of sound in air = 340340 m/s)

Explanation

Solution

There are two questions in the problem, first we have to find the time taken for the stone to drop from the top of the building to the pond and next is the time taken by the sound of the splash to reach the top of the building, where the sound is to be heard. Apply the basic formula of motion under gravity and find it.

Complete step by step answer:
Let us find the time taken for the stone to fall from the top of the building to the pond at first. Let the top of the building be A and its bottom, where the pond is B. Let time taken to travel from A to B be t1t_1 . Let the height of AB be H.

Now from the formula of motion under gravity:
H=ut+12gt2H = ut + \dfrac{1}{2}g{t^2}
Where uu = initial speed, tt = time taken and aa = acceleration. Here a=ga = g, as an item is falling under gravity.Here the item is freely falling, thus u=0u = 0 .
Thus by applying the formula we get 525=12×g×t12525 = \dfrac{1}{2} \times g \times t{1^2}.
Thus t1=10.35st_1 = 10.35\,s.

Now we have to calculate the speed for the sound to travel from the pond to 525m525\,m above, at the top of the building. Thus, knowing the distance, we can easily calculate time by using the formula of distance=speed×time\text{distance} = \text{speed} \times \text{time}.
Thus, the time t2t_2 taken for sound to travel from B to A = 525340=1.55s\dfrac{{525}}{{340}} = 1.55\,s
Hence total time = t1+t2=11.9st_1 + t_2 = 11.9s.

Hence,the sound of the splash can be heard at 11.9 s.

Note: Note that at the first case we consider the influence of gravity for the free fall of the stone, but in the 2nd2^{nd} case we do not do so, as in every case generally we consider the speed of sound to be a constant , whose value is given here as 340340 m/s.