SATHEE: Kinematics Question 208

Kinematics Question 208

Question: A body falling for 2 seconds covers a distance $S$ equal to that covered in next second. Taking $g = 10 m / s^{2}, S =$ [EAMCET (Engg.) 1995]

Options:

A) 30 m

B) 10 m

C) 60 m

D) 20 m

Show Answer

Answer:

Correct Answer: A

Solution:

If u it’s the initial velocity then distance covered by it in 2 sec $S = u t + \frac{1}{2} a t^{2} = u \times 2 + \frac{1}{2} \times 10 \times 4 = 2 u + 20$ -(i)

Now distance covered by it in 3rd sec $S_{3^{r d}} = u + \frac{g}{2} (2 \times 3 - 1) 10 = u + 25$ -(ii)

From(i) and (ii), $2 u + 20 = u + 25 \Rightarrow u = 5$
$∴$ $S = 2 \times 5 + 20 = 30 m$

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Kinematics Question 208

Question: A body falling for 2 seconds covers a distance S equal to that covered in next second. Taking g=10m/s2,S= [EAMCET (Engg.) 1995]

Options:

Answer:

Solution:

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Question: A body falling for 2 seconds covers a distance $S$ equal to that covered in next second. Taking $g = 10 m / s^{2}, S =$ [EAMCET (Engg.) 1995]