Question

A stone is thrown horizontally. In 0.5 second after the stone began to move, the magnitude of its velocity was 1.5 times its initial velocity. Find the initial velocity of the stone.

(Pls help me with the dig of this question )

Answer 1

So, as the body was projected horizontally,

Ux = v

Uy=0

(Assuming the initial velocity to be v)

Consider x and y to be standard coordinate axes

Now, there is no acceleration in the x direction, so x component of velocity will remain same throughout the flight

But y component has force of gravity so, it will result in velocity increase

Now,

After 0.5 sec,

Ux=v

For y component,

v=u+ at

a=-10m/s²

So,

v= 0 -10 ×0.5

v=-5 (-ve due to considering the sign conventions)

Vy= 5m/s 

Now,

the velocity after 0.5 sec is 1.5 times the initial velocity

So,

Vnet=1.5v

So,

Vnet=√(Vx² +Vy²)

(Law of vector addition)

1.5v=√(25+ v²)

2.25v² - v² = 25

v²=20

v=4.47 m/s

So, the initial velocity was 4.47 m/s

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