the distance travelled by a body falling freely from in the 1st, 2nd and 3rd seconds, are in the ratio
a)1:2:3 b)1:3:5 c)1:4:9 d)3:2:1

Question

the distance travelled by a body falling freely from in the 1st, 2nd and 3rd seconds, are in the ratio 
a)1:2:3   b)1:3:5  c)1:4:9      d)3:2:1

luffingsails · Answer

adhira,

A free-falling body will accelerate. This should help you answer the question. Only one of the answers indicates this behavior.

anonymous · Answer

What do you mean by 1st, 2nd and 3rd second?
Can you explain your question a little more.

anonymous · Answer

pls can u explain that i haven't the answer key of this question upsilon

anonymous · Answer

but there u=0

luffingsails · Answer

given gravity =32 ft/sec/sec

d = 16 t^2

So, distances for seconds 1, 2, and 3 = 16, 64, and 144 respectively. The ratio that represents these is 1:4:9.

anonymous · Answer

yeah i have the same ans 1:4:9 
it's my exam question today

anonymous · Answer

$$v = u +gt $$ 
Now at $t = 0 $, $u=0$.
$$v = gt \implies v = 10$$
$$s = ut + \frac{1}{2}*g*t^2$$
$$s _1= 5 m $$ 
For the second time interval the final velocity of 1st interval will be initial velocity 
$$s_2 = 10 + 5$$
$$s _2= 15$$
Now final velocity for second interval $v = 10 + gt$, $v=20$ 
$$s_3 = 20 + 5$$

$$s_1:s_2:s_3 = 1:3:5$$

anonymous · Answer

thanks thanks thanks thanks UPSILON U DONE  A VERY VERY GOOD JOB

anonymous · Answer

Naruto San ...Sensei here

anonymous · Answer

Yea I am with upsilon