Kinematics Question 230
Question: From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s. The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (Take $ g=10m/s^{2} $ ) [AIIMS 2000; CBSE PMT 2002]
Options:
A) 5 : 7
B) 7 : 5
C) 3 : 6
D) 6 : 3
Show Answer
Answer:
Correct Answer: B
Solution:
$ {S _{3^{rd}}}=10+\frac{10}{2}(2\times 3-1)=35\ m $
$ {S _{2^{nd}}}=10+\frac{10}{2}(2\times 2-1)=25\text{ m} $
therefore $ \frac{{S _{3^{rd}}}}{{S _{2^{nd}}}}=\frac{7}{5} $
BETA
