A stone is allowed to fall from the top of a tower 100 m high

Solution:

For stone moving downward, acceleration due to gravity, $\mathrm{g}=10 \mathrm{~ms}^{-2}$

initial velocity (u) = 0 ; distance, s = 100 – x ;

time, t = ?

$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$, or $(100-\mathrm{x})=0 \times \mathrm{t}+\frac{1}{2} \times 10 \mathrm{t}^{2}$

$\Rightarrow 100-x=5 t^{2}$ ....(1)

For the stone moving vertically upwards,

initial velocity, $\mathrm{u}=25 \mathrm{~ms}^{-1} ;$ time $(\mathrm{t})=? ;$ acceleration due to gravity, $\mathrm{g}=-10 \mathrm{~ms}^{-1}$

[In upward direction, g is taken –ve]



or $\quad x=25 \times t+\frac{1}{2} \times\left(-10 t^{2}\right)$

$\Rightarrow \quad x=25 t-5 t^{2}$ ...(2)

Substituting the value of x from (2) in (1) we get,

$100-\left(25 \mathrm{t}-5 \mathrm{t}^{2}\right)=5 \mathrm{t}^{2}$

$100-25 t+5 t^{2}=5 t^{2}$

$25 \mathrm{t}=100$ or $\mathrm{t}=4 \mathrm{~s}$

Put the value of t in (1),

(1) $\Rightarrow \quad 100-x=5(4)^{2}$

$\Rightarrow \quad 100-x=80$ or $x=20 \mathrm{~m}$

$\therefore$ The stones will meet at a height of $20 \mathrm{~m}$ from ground after $4 \mathrm{~s}$.