Question.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stoneis projected vertically upwards from the ground with a velocity of 25 ms–1. Calculate when and where the two stones will meet. (Take g = 10 m/s2)
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stoneis projected vertically upwards from the ground with a velocity of 25 ms–1. Calculate when and where the two stones will meet. (Take g = 10 m/s2)
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 $4 \mathrm{~s}$.
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 $4 \mathrm{~s}$.
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