A body of mass m is projected horizontally with a velocity v from the top of a tower of height h ; and it reaches the ground at a distance x from the foot of the tower. If a second body of mass 2m is projected horizontally from the top of a tower of height 2h, it reaches the ground at a distance 2x from the foot of the tower. The horizontal velocity of the second body is:

Question

A body of mass m is projected horizontally with a velocity v from the top of a tower of height h ; and it reaches the ground at a distance x from the foot of the tower. If a second body of mass 2m is projected horizontally from the top of a tower of height 2h, it reaches the ground at a distance 2x from the foot of the tower. The horizontal velocity of the second body is: (A) v (B) 2v (C) √2v (D) v/2

Tardigrade Admin · Accepted Answer

For the first body, h=(1/2) gt2 (i) and x=vt (ii) From equation (i) and (ii) h=(1/2)g. ((x2/v2)) (iii) Dividing equation (iii) by equation (iv) we get (1/2)= (x2/v2) × (v'2/4x2) or v' =√2v

$v$

$2 v$

$2 v$

$v /2$

For the first body, $h = \frac{1}{2} g t^{2} (i)$

and $x = v t (ii)$

From equation (i) and (ii)

$h = \frac{1}{2} g . (\frac{x ^{2}}{v ^{2}}) (iii)$

Dividing equation (iii) by equation (iv) we get

$\frac{1}{2} = \frac{x ^{2}}{v ^{2}} \times \frac{v ^{' 2}}{4 x ^{2}}$ or $v^{'} = 2 v$

v

2v

2v​

v/2

For the first body, h=21​gt2(i) and x=vt(ii) From equation (i) and (ii) h=21​g.(v2x2​)(iii) Dividing equation (iii) by equation (iv) we get 21​=v2x2​×4x2v′2​ or v′=2v​

$v$

$2 v$

$2 v$

$v /2$

For the first body, $h = \frac{1}{2} g t^{2} (i)$

and $x = v t (ii)$

From equation (i) and (ii)

$h = \frac{1}{2} g . (\frac{x ^{2}}{v ^{2}}) (iii)$

Dividing equation (iii) by equation (iv) we get

$\frac{1}{2} = \frac{x ^{2}}{v ^{2}} \times \frac{v ^{' 2}}{4 x ^{2}}$ or $v^{'} = 2 v$