Question

A man standing on the roof of a house of height h throws one particle vertically downwards and another particle

horizontally with the same velocity u. The ratio of their velocity when they reach the earth's surface will be –

• A. $\sqrt{2gh + u^{2}}:u$
• B. 1 : 2
• C. 1 : 1
• D. $\sqrt{2gh + u^{2}}:\sqrt{2gh}$

Answer 1

Answer: (C)

1 : 1

Answer 2

From mechanical energy conservation principle

$\frac { 1 } { 2 }$ m (v² –u²) = mgH

\ v₁ = v₂ or $\frac { v _ { 1 } } { v _ { 2 } }$ = 1