Question

A stone thrown upwards with a velocity u reaches upto a height h . If the initial velocity is 2u the height attained would be​

Answer 1

Answer:-

H = 4h

Given:-

Case :- 1

Initial velocity = u

Height covered by Stone = h

g = -10 m/s²

Solution :-

Since, the object is thrown upward so the acceleration due to gravity will be in negative direction.

Now,

The maximum height reached by stone is given by :-

$\boxed{\sf{h_{max} = \dfrac</p><p>{u^2}{2g}}}$

→$h = \dfrac{u^2}{2g}$

Case :- 2

If it's initial velocity will be 2u.

Let the height cover will be H metre →$H = \dfrac{(2u)^2}{2g}$

→$H = \dfrac{4u^2}{2g}$

→$H = 4\dfrac{u^2}{2g}$

→$H = 4 h$

hence, the height covered by stone is four times than previous height.