SATHEE: Kinematics Question 180

Kinematics Question 180

Question: A stone is dropped into water from a bridge $44.1$

$m$ above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone

Options:

A) $12.25 m / s$

B) $14.75 m / s$

C) $16.23 m / s$

D) $17.15 m / s$

Show Answer

Answer:

Correct Answer: A

Solution:

Time taken by first stone to reach the water surface from the bridge be t,

then $h = u t + \frac{1}{2} g t^{2} \Rightarrow 44.1 = 0 \times t + \frac{1}{2} \times 9.8 t^{2}$

$t = \sqrt{\frac{2 \times 44.1}{9.8}} = 3 s e c$

Second stone is thrown 1 sec later and both strikes simultaneously.

this means that the time left for second stone $= 3 - 1 = 2 s e c$ Hence $44.1 = u \times 2 + \frac{1}{2} 9.8 {(2)}^{2}$
$\Rightarrow 44.1 - 19.6 = 2 u \Rightarrow u = 12.25 m / s$

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Kinematics Question 180

Question: A stone is dropped into water from a bridge 44.1

Options:

Answer:

Solution:

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Question: A stone is dropped into water from a bridge $44.1$