Question

A body is released from rest from the top of a
tower of height 3H. The ratio of times it takes
to fall through equal height H is :. please answer
...​

Answer 1

Explanation:

your question is in complete..

Q)

A particle is released from rest from a tower of height 3h, the ratio of time of fall for equal heights h ie t1:t2:t3 is

answer -

h= 1/2gt1^2

2h= 1/2g(t1+t2)^2

3h= 1/2(t1+t2+t3)^2

t1:(t1+t2):(t1+t2+t3) = 1:✓2:✓3

or

t1:t2:t3= 1:(✓2-1):(✓3-✓2)

hope it helps.......

Answer 2

Answer:

h=12h=12gt21gt12

2h=122h=12g(t1+t2)2g(t1+t2)2

3h=123h=12g(t1+t2+t3)2g(t1+t2+t3)2

t1:(t1+t2):(t1+t2+t3)=1:2–√:3–√t1:(t1+t2):(t1+t2+t3)=1:2:3

or t1:t2:t3=1:(2–√−1):(3–√−2–√)t1:t2:t3=1:(2−1):(3−2)

Hence d is the correct answer.