A ball is dropped from the top of a tower of height $100 \mathrm{~m}$ and at the same time another ball is projected vertically upwards from ground with a velocity $25 \mathrm{~ms}^{-1}$. Then what will be the distance (in $\mathrm{m}$ ) from the top of the tower, at which the two balls meet?

Question

Aicharya · Accepted Answer

Let the two balls Pand $Q$ meet at height $x \mathrm{~m}$ from the ground after time $t \mathrm{~s}$ from the start. We have to find distance, $B C=(100-x)$

For ball P $S=x \mathrm{~m}, u=25 \mathrm{~m} \mathrm{~s}^{-1}, a=-g$
From $S=u t+\frac{1}{2} a t^{2}$
$x=25 t-\frac{1}{2} g t^{2} \quad$..........(i)
For ball $Q$ $S=(100-x) \mathrm{m}, u=0, a=g$
$\therefore 100-x=0+\frac{1}{2} g t^{2} \quad \ldots \ldots \ldots$ (ii)
Adding eqns. (i) and (ii), we get $100=25 t$ or $t=4 \mathrm{~s}$
From eqn. (i), $x=25 \times 4-\frac{1}{2} \times 9.8 \times(4)^{2}=21.6 \mathrm{~m}$
Hence distance from the top of the tower $=(100-x) \mathrm{m}=(100-21.6 \mathrm{~m})=78.4 \mathrm{~m}$ .

A ball is dropped from the top of a tower of height $100 \mathrm{~m}$ and at the same time another ball is…

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