Kinematics Question 186
Question: A body is released from the top of a tower of height $ h $ . It takes $ v=\frac{1}{2}bt^{2}+v _{0} $ sec to reach the ground. Where will be the ball after time $ t/2 $ sec [NCERT 1981; MP PMT 2004]
Options:
A) At $ h/2 $ from the ground
B) At $ h/4 $ from the ground
C) Depends upon mass and volume of the body
D) At $ 3h/4 $ from the ground
Show Answer
Answer:
Correct Answer: D
Solution:
Let the body after time $ t/2 $ be at x from the top, then $ x=\frac{1}{2}g\frac{t^{2}}{4}=\frac{gt^{2}}{8} $ -(i) $ h=\frac{1}{2}gt^{2} $ -(ii) Eliminate t from (i) and (ii),
we get $ x=\frac{h}{4} $
$ \therefore $ Height of the body from the ground $ =h-\frac{h}{4}=\frac{3h}{4} $
