water drops fall at regular intervals from a roof , at an instant when a drop is about to leave the roof, the separations between 3 successive drops below the roof are in what ratio?
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Answered by
18
The answer is 5:3:1
Let the interval be t. Then the first drop of the three has left t sec before, second has 2t sec and the third one at 3t sec. Then the distances covered by them respectively are
0(Drop about to fall), 1/2(t2) (I drop) , 2t2 (II drop), 9/2 t2 (III drop)
The distances b/w them r in ratio (5/2)t2 : (3/2)t2 : (1/2)t2= 5:3:1
Let the interval be t. Then the first drop of the three has left t sec before, second has 2t sec and the third one at 3t sec. Then the distances covered by them respectively are
0(Drop about to fall), 1/2(t2) (I drop) , 2t2 (II drop), 9/2 t2 (III drop)
The distances b/w them r in ratio (5/2)t2 : (3/2)t2 : (1/2)t2= 5:3:1
Answered by
0
Answer:
the required ratio will be ...
1:3:5
hope it helps you ✨
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