Question

show your guts by solving question no. 13​

Answer 1

Let the depth of the well be s. This s travelled by the stone to reach the surface of water is related to the time t1 by the second kinematic equation,

s = ut + (at²) / 2

Here, u = 0 and a = g = 9.8 m s^(-2). Thus,

s = 9.8 (t1)² / 2

s = 4.9 (t1)² → (1)

Well, the same distance is travelled by the sound of the splash, and is given by the standard equation,

Displacement = Speed × Time

Here time is t2 and speed is the speed of sound, i.e., 330 m s^(-1). So,

s = 330 (t2) → (2)

But given,

(t1) + (t2) = 2.5

t2 = 2.5 - (t1)

Thus, (2) becomes,

s = 330 [2.5 - (t1)]

s = 825 - 330(t1) → (3)

Now, equating (1) and (3),

4.9(t1)² = 825 - 330(t1)

4.9(t1)² + 330(t1) - 825 = 0

t1 = (- 330 + √(108900 + 16170)) / 9.8

[Time can't be negative, hence positive square root is taken.]

t1 = 2.41 s

Hence (1) becomes,

s = 4.9 (2.41)²

s = 28.6 m