Question

A stone is thrown horizontally with velocity g m/s from the top of a tower of height g meter.The velocity with which it hits the ground

Answer 1

Answer:

√3 g

Explanation:

horizontal component of velocity never changes = g m/s

v^2 = u^2 + 2as

by vector addition rule they r perpendicular so resultant velocity is

√3g

Answer 2

Answer:

The velocity with which the stone hits the ground is $\sqrt{3} g$ m/s

Explanation:

Since the stone is thrown horizontally, the vertical component of the velocity will be zero, it will have only horizontal component

Horizontal component of the velocity = g m/s

or, $v_x=g$ m/s

Since there is no acceleration in the horizontal direction, the horizontal component of the velocity will be constant throughout its trajectory

Height of the tower h = g m

using the third equation of motion in the vertical direction

if final velocity is $v_y'$ then

$(v_y')^2=0^2+2g\times g$

or, $v_y'=\sqrt{2} g$ m/s

Final horizontal component of the velocity

$v_x'=g$ m/s

Therefore, the velocity with which the stone hits the ground

$=\sqrt{(v_y')^2+(v_x')^2}$

$=\sqrt{(\sqrt{2}g )^2+(g)^2}$

$=\sqrt{3} g$ m/s