Question

A stone is thrown horizontally with velocity g ms from the top of a tower of height g
metre. The velocity with which it hits the ground is g. Then the value of N is (in ms")
2​

Answer 1

Explanation:

The velocity with which the stone hits the ground is \sqrt{3} g3g 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=gvx=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'vy′ then

(v_y')^2=0^2+2g\times g(vy′)2=02+2g×g

or, v_y'=\sqrt{2} gvy′=2g m/s

Final horizontal component of the velocity

v_x'=gvx′=g m/s

Therefore, the velocity with which the stone hits the ground

=\sqrt{(v_y')^2+(v_x')^2}=(vy′)2+(vx′)2

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

=\sqrt{3} g=3g m/s

Answer 2

Answer:

The velocity with which the stone hits the ground is \sqrt{3} g3g 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=gvx=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'vy′ then

(v_y')^2=0^2+2g\times g(vy′)2=02+2g×g

or, v_y'=\sqrt{2} gvy′=2g m/s

Final horizontal component of the velocity

v_x'=gvx′=g m/s

Therefore, the velocity with which the stone hits the ground

=\sqrt{(v_y')^2+(v_x')^2}=(vy′)2+(vx′)2

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

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

Explanation: