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 velocities when they reach the earth's surface will be

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 velocities when they reach the earth's surface will be (A) √2 g h+u2: u (B) 1: 2 (C) 1: 1 (D) √2 g h+u2: √2 g h

Tardigrade Admin · Accepted Answer

When particle is thrown in vertical downward direction with velocity u, then the final velocity at the ground level is v2 =u2+2 g h ∴ v =√u2+2 g h Another particle is thrown horizontally with same velocity then at the surface of earth.

Horizontal component of velocity vx = u Therefore, resultant velocity, v = √u2 + 2gh For both the particles, final velocities when they reach the earth's surface are equal.

Q. 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 velocities when they reach the earth's surface will be

$2 g h + u^{2} : u$

$1 : 2$

$1 : 1$

$2 g h + u^{2} : 2 g h$

Q. 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 velocities when they reach the earth's surface will be

2gh+u2​:u

1:2

1:1

2gh+u2​:2gh​

Q. 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 velocities when they reach the earth's surface will be

$2 g h + u^{2} : u$

$1 : 2$

$1 : 1$

$2 g h + u^{2} : 2 g h$