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 2 m 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- (A) 2v (B)√2v (C) v/2 (D) v/√2

Answer 1

Explanation:

Time taken to cover height h=

g

2h

Distance covered in same time x=vt=v

g

2h

Time taken to cover height 2h=

g

4h

Distance covered in same time 2x=vt=v

g

4h

⇒2V

g

2h

=v

g

4h

⇒V=

2

v

Answer 2

Answer:$\sqrt{2v}$

Explanation:Time taken to cover height h=v  $\sqrt{2h/g\\}$

Distance covered in same time x=vt=v$\sqrt{2h/g}$

Time taken to cover height 2h=$\sqrt{4h/g}$

=2V$\sqrt{2h/g\\}$=v $\sqrt{4h/g}$

v=$\sqrt{2v}$

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