jee mains related questions...plz give the correct ans...if u dont know plz dont reply just for the sake of points
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jee mains related questions...plz give the correct ans...if u dont know plz dont reply just for the sake of points
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Given :-
A ball is thrown vertically upwards with velocity "v" from the top of a tower and it reaches the ground with speed "4v" and height of the tower is [tex] \dfrac{nv²}{2g} [/tex] .
To Find :-
The Value of "n"
Used Concepts :-
Solution :-
First see the attachment for better understanding of the question ;
From the diagram it is clear that ;
Initial velocity of ball = u = v
Final Velocity of ball = m = 4v
g = 9.8 m/s²
Height of the tower = The distance traveled by ball = s = [tex] \dfrac{nv²}{2g} [/tex]
By Newton's Law of motion ;
=> m² - u² = 2gs
=> ( 4v )² - ( v )² = 2 × [tex] \dfrac{nv²}{2g} × g [/tex]
=> 16v² - v² = nv²
=> 15v² = nv²
=> n = [tex] \dfrac{15v²}{v²} [/tex]
=> n = 15
Henceforth , Value of n is 15 .