Physics, asked by anuprajranchi, 8 months ago

From a leak tap, water droplets are constantly falling. It is 11.25 m high. When 1st drop

reaches to ground, 4th drop comes out. If the distance between 2nd and 3rd drop at this

instant is x meter then calculate the value 100x.

(g = 10 m/s2

)​

Answers

Answered by Anonymous
59

Given:

Height of the tap from ground (h) = 11.25 m

When  \sf 1^{st} drop reaches ground  \sf 4^{th} drop comes out of tap and distance between  \sf 2^{nd} and  \sf 3^{rd} drop is x meter.

Accelration due to gravity (g) = 10 m/s²

To Find:

Value of 100x

Answer:

By using  \sf 2^{nd} equation of motion we can find total time required by  \sf 1^{st} drop to reach the ground from total time we can calculate the time interval between the subsequent drops.

Initial velocity of drop will be 0 m/s. So,

 \bf \implies h = ut +  \dfrac{1}{2} g {t}^{2}  \\  \\  \rm \implies 11.25 =  \dfrac{1}{2}  \times 10 {t}^{2}  \\  \\ \rm \implies  {t}^{2}  =  \dfrac{11.25 \times 2}{10}  \\  \\ \rm \implies  {t}^{2}  = 2.25 \\  \\ \rm \implies t =  \sqrt{2.25}  \\  \\ \rm \implies t = 1.5 \: s

From this we get that time interval between subsequent drop is 0.5 s

When the  \sf 1^{st} drop reaches the ground;

Distance covered by  \sf 2^{nd} drop will be:

 \rm \implies h_2 = 0 + \dfrac{1}{2} \times 10 \times 1^2 \\ \\  \rm \implies h_2 =  \dfrac{1}{2} \times 10 \\ \\ \rm \implies h_2 =  5 \ m

Distance covered by  \sf 3^{rd} drop will be:

 \rm \implies h_3 = 0 + \dfrac{1}{2} \times 10 \times (0.5)^2 \\ \\  \rm \implies h_3 =  \dfrac{1}{2} \times 10 \times 0.25 \\ \\ \rm \implies h_3 =  1.25 \ m

Thus,

Distance between  \sf 2^{nd} &  \sf 3^{rd} drop:

 \rm \implies x = h_2 - h_3 \\  \\ \rm \implies x = 5  - 1.25 \\  \\ \rm \implies x = 3.75 \: m

 \therefore  \boxed{\mathfrak{100x = 375 \ m}}

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