CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
b, c & d are correct
loader
C
a & c are correct
loader
D
a, b, c & d are correct
loader

Solution

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

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image