Question

the velocity acquired by the body when it falls through a height h is v. If it further falls through a distance x where (x<<h) then find the increase in velocity?​

Answer 1

Answer:

$(A) \frac{vx}{2h}$

Explanation:

1) We have,

Velocity acquired by a body  when it falls through height h,

$=>v=\sqrt{2gh}$

2) Now, it further falls through a height x (x<<h) ,

then

Using Newtons equation of motion,

$v_1^2-v^2=2gx\\ \\=>v_1^2=v^2+2gx\\ \\=>v_1=\sqrt{v^2+2gx}$

3) Now,

$v_1-v=\sqrt{v^2+2gx}-v\\ \\=v\sqrt{1+\frac{2gx}{v^2}}-v\\ \\=v\sqrt{1+\frac{2gx}{2gh}}-v\\ \\=v(\sqrt{1+\frac{x}{h}}-1)\\ \\=v((1+\frac{x}{h})^{1/2}-1)\\ \\ \approx v(1+\frac{x}{2h}-1)\:\:\:(\because x<<h)\\ \\=\frac{vx}{2h}$

Hence, increase in velocity is approximately vx/2h.