Question

a ring rolls down from the top of an inclined plane pf vertical height h. It will reach the ground after a time
1) 1/sin theta root 4h/g
2) 1/sin theta root 2h/g
3) 1/sin theta root 3h/g
4)1/sin theta root 14h/5g

Answer 1

Answer:

2

1/sin theta root 2h/g

Answer 2

Given:

The top of an inclined plane vertical height h.

To find:

Time taken.

Explanation:

Acceleration due to gravity along inclined,

$g^{'}=gcos(90^{0}-0)=gsin$∅

Time taken $t=\sqrt{\frac{2s}{g} }$

$=\sqrt{\frac{2l}{gsin} }$

$t=\sqrt{\frac{2}{gsin}*\frac{h}{sin} }$

$=\frac{1}{sin}$∅$\sqrt{\frac{2h}{g} }$

Answer:

Therefore, The time taken in the $1/sin theta root 2h/g$.