Question

A tank is filled with water and two holes A and B are made in it. For getting same range , ratio of h/h' is??

see the attachment​

Answer 1

Range(A)=2hg×2g(x+h′ ) Range(B)=2(h+x) g×2gh′ h(x+h′ )=h′ (x+h)hx+hh′ =hh′ +h′ x h=h′  hh′ =1Range\left( A \right) =\sqrt { \cfrac { 2h }{ g } \times 2g\left( x+{ h }^{ \prime  } \right)  } \\ Range\left( B \right) =\sqrt { \cfrac { 2\left( h+x \right)  }{ g } \times 2g{ h }^{ \prime  } } \\ h\left( x+{ h }^{ \prime  } \right) ={ h }^{ \prime  }\left( x+h \right) \\ hx+h{ h }^{ \prime  }=h{ h }^{ \prime  }+{ h }^{ \prime  }x\\  \quad \quad \quad h={ h }^{ \prime  }\\ \quad  \quad \cfrac { h }{ { h }^{ \prime  } } =1Range(A)=g2h​×2g(x+h′ ) (all in root)

Range(B)=g2(h+x) ​×2gh′ (all in root)

h(x+h′ )=h′ (x+h)hx+hh′ =hh′ +h′ x h=h′  h′ h​=1(all in root)

Answer 2

Answer:

Explanation:

for height h'

velocity at hole is $v'=\sqrt{2gh'}$

time taken to reach ground is

$t'=\sqrt{\frac{2(x+h)}{g}}$

where x is the distance between h and h'

Range for $h'=v't'=\sqrt{2gh'}\times \sqrt{\frac{2(x+h)}{g}}$

For height h

exit velocity is

$v=\sqrt{2gh}$

time taken to cover h height

$t=\sqrt{\frac{2h}{g}}$

Range for $h=v\times t$

Range$=\sqrt{2gh}\times \sqrt{\frac{2h}{g}}$

Range for both the height is same therefore

$\sqrt{2gh'}\times \sqrt{\frac{2(x+h)}{g}}=\sqrt{2gh}\times \sqrt{\frac{2h}{g}}$

$h'(h+x)=(h'+x)h$

$h'x+hh'=hh'+hx$

$x(h-h')=0$

thus $h=h'$

ratio of $\frac{h}{h'}=1$