Question

A stone is dropped into water from a bridge 44.1m above the water. Another stone is thrown vertically downward one second later. Both strike the water simulataneously, then the initial speed
of the second stone is?

Answer 1

Answer:

Step 1: Time taken (t) by stone 1

(Taking downward positive)

Since acceleration is constant, therefore applying equation of motion

h=ut+

2

1

gt

2

⇒ 44.1=0+

2

1

(9.8)t

2

⇒ t=3s

Step 2: Initial Speed Calculation

Let the initial speed of stone 2 be u

Both the stones strike the water simultaneously but the stone 2 is thrown 1s later which means it takes 1s less as compared to stone 1 to cover the same distance.

Therefore, time taken by stone 2, t

1

=t−1 =3−1 s=2 s

Since acceleration is constant

∴ Applying equation of motion (Taking downward positive)

h=ut

1

+

2

1

gt

1

2

⇒44.1=u×2+

2

1

×9.8×(2)

2

⇒u=12.25 m/s

Hence option A is correct.

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