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From the top of a tower of height h a body of mass m is projected in the horizontal direction with the velocity V it falls on the ground at a distance x from the tower if the body of mass 2 m is projected from the top of another Tower of height 2 h in the horizontal direction so that it falls on the ground at the distance to X from the tower the horizontal velocity of the second body is?

From the top of a tower of height h a body of mass m is projected in the horizontal direction with the velocity V it falls on the ground at a distance x from the tower if the body of mass 2 m is projected from the top of another Tower of height 2 h in the horizontal direction so that it falls on the ground at the distance to X from the tower the horizontal velocity of the second body is?

Grade:11

1 Answers

Laksh
23 Points
6 years ago
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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