Question

Consider the following statements regarding the K.E of a freely falling body :
a) is directly proportional to height of its fall
b) is inversely proportional to height of its fall
c) is directly proportional to square of time of its fall
d) is inversely proportionl to square of time of its fall

• A. a, b & d are correct
• B. b, c & d are correct
• C. a & c are correct
• D. a, b, c & d are correct

Answer 1

The correct option is C a & c are correct

$M.E=K.E+P.E$
$0=K.E-mgh$

So, $K.E=mgh$

KE is proportional to h
and $h=-\frac{1}{2}g{t}^2$ as for a freely falling body initial velocity is zero.
$\Rightarrow K.E=-\frac{m{g}^2{t}^2}{2}$
So KE is proportional to $t^2$.