A stone is allowed to fall from the top of a tower $100\ m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25\ m/s$. Calculate when and where the two stones will meet.

For stone A: When the stone from the top of the tower is thrown,

Initial velocity $u_A=0$

Distance traveled $=x$

Time taken $=t$

Therefore,

Height, $h=(100 – x)$

Here, $g=10\ m/s^2$     [As the stone is falling down]

We know that,

$h=ut+\frac{1}{2}gt^2$

On substituting the values in the above equation

$(100-x)=0\times t+\frac{1}{2}\times 10\times t^2$

Or $(100-x)=5t^2$ .......$(i)$

For stone B: for stone projected vertically upwards:

Height, $h=x$

Initial velocity, $u_B=25\ m/s$

Time, $t=?$

$g=-10\ m/s^2$       [Because the stone is going up]

On using, $s=ut+\frac{1}{2}gt^2$

Or $x=25t+\frac{1}{2}(-10)t^2$

Or $x=25t-5t^2$

On adding the above equations We get

$100=25t$

or $t=\frac{100}{25}=4\ s$

After $4\ second$, the two stones A and B will meet.

On putting the value $t=4\ s$ in $(i)$

$100-x=5t^2$

Or $x=100-5t^2$

Or $x=100-5\times4^2$

Or $x=100-80$

Or $x=20\ m$

So, after $4\ seconds$, two stones meet at a distance of $20\ m$ from the ground.

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Updated on: 10-Oct-2022

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