Question

From the top of a tower of height h a body of mass m is projected in the horizontal direction with a

velocity v, it falls on the ground at a distance x from the tower. If a body of mass 2m is projected from

the top of another tower of height 2h in the horizontal direction so that it falls on the ground at a

distance 2x from the tower, the horizontal velocity of the second body is​

Answer 1

Answer:

Explanation:

Case 1

velocity=v range=x

height of the tower =h

Therefore using the horizontal equation of range,

R=u{\sqrt{(2h)/g)}}

therefore x=v{\sqrt{(2h)/g)}} --------- 1

Case 2

velocity=2v range=2x

height of the tower =2h

Therefore using the horizontal equation of range,

R=u{\sqrt{(2h)/g)}}

2x=vnew{\sqrt{(4h)/g)}} --------- 2

eq 1/eq 2

\frac{x}{2x}=\frac{v{\sqrt{(2h)/g)}} }{vnew{\sqrt{(4h)/g)}} }

1=\frac{v\sqrt{2}}{vnew}

therefore vnew =v\sqrt{2}

In the above question we neglect the mass because ‘g’ is independent of the mass

Hope this helps!!!!

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