Question

A stone is projected vertically upwards from a point
the ground with velocity of soms Two
seconds later another stone is fusfected from the
same point with the same velocity When and
where they meet? 19 -
= 10 m/s²)​

Answer 1

Answer:

Let's take the first stone as (A) and the second stone as ( B )

let's take the time taken by stone (A)

as (t)

and time taken by stone (B) as (t-2)

to find the time taken by the stones to meet

we have get separate equations for stone A and stone B...

equation for stone A

$s = ut + \frac{1}{2}a {t}^{2} \\ h = ut + \frac{1}{2} ( - g) {t}^{2} \\ h = ut - \frac{1}{2} g {t}^{2}$

equation for stone B

$s = ut + \frac{1}{2} a {t}^{2} \\ h = u(t - 2 )+ \frac{1}{2} ( - g) (t - 2)^{2} \\ h= ut - 2u - \frac{1}{2} (t - 2)^{2}$

and then substitute the equation 1 to 2and equation

$ut -\frac{g}{2}{t}^{2} =ut - 2u - \frac{g}{2}{t}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: -2g+2gt\\ \frac{2u}{2g} + \frac{2g}{2g} = \frac{2gt}{2g} \\ t = \frac{u}{g} + 1$