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Q. From the 9 m high tap of water, droplets are falling regularly such that when 4^th drop starts, then 1^st drop strikes the ground. The position of 3^rd drop from the ground is (g = 10 m/s²)

a. 8 m
b. 1 m
c. 3 m
d. 4 m

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Solution

Dear Student,

$l e t t h e i r i s t i m e g a p o f t b e t w e e n t h e d r o p o f d r o p l e t s . s o t h e d r o p 1 r e a c h a f t e r t i m e 3 t 9 = \frac{10 * 9 t^{2}}{2} t^{2} = \frac{1}{5} p o s i t i o n o f d r o p s = \frac{10 * t^{2}}{2} = 1 m f r o m t h e t a p . p o s i t i o n o f 3 r d d r o p f r o m t h e g r o u n d = 9 - 1 = 8 m R e g a r d s$

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Solve this: Q. From the 9 m high tap of water, droplets are falling regularly such that when 4th drop starts, then 1st drop strikes the ground. The position of 3rd drop from the ground is (g = 10 m/s2) a. 8 m b. 1 m c. 3 m d. 4 m

Dear Student, let their is time gap of t between the drop of droplets .so the drop 1 reach after time 3t9=10*9t22t2=15 position of drop s=10*t22=1m from the tap .position of 3rd drop from the ground =9-1= 8 mRegards

Solve this:

Q. From the 9 m high tap of water, droplets are falling regularly such that when 4^th drop starts, then 1^st drop strikes the ground. The position of 3^rd drop from the ground is (g = 10 m/s²)

a. 8 m
b. 1 m
c. 3 m
d. 4 m