Question

A stone is thrown horizontally from an elevated point. After 0.5 seconds, the magnitude of its velocity is 1.5 times the magnitude of its initial velocity. Find the initial speed of stone.​

Answer 1

Answer:

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

HOPE THAT IT WAS HELPFUL!!!!

MARK IT THE BRAINLIEST IF IT REALLY WAS!!!!