A ball is thrown downwards with a speed of \(20 \mathrm{~ms}^{-1}\) from the top of a building \(150 \mathrm{~m}\) high and simultaneously another ball is thrown vertically upwards with a speed of \(30 \mathrm{~ms}^{-1}\) from the foot of the building. Find the time after which both the balls will meet. \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

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Aicharya · Accepted Answer

Method-1: Let the first ball move down distance \(S_{1}\) and second ball moves up a distance \(S_{2}\) before they meet. Let us take downward direction as positive.
Then \(S_{1}=20 t+5 t^{2}\)
\(S_{2}=30 t-5 t^{2}\)
But, \(\quad S_{1}+S_{2}=150\)
\(\Rightarrow \quad 150=50 t\)
\(\Rightarrow \quad t=3 \mathrm{~s}\)
Method-2: Let us solve this problem by using the method of relative velocity.
Relative acceleration of both is zero since both have same acceleration in downward direction.
\(\vec{a}_{A B}=\vec{a}_{A}=-\vec{a}_{B}=g-g=0\)
Initial relative velocity, \(\vec{v}_{B A}=30-(-20)=50 \mathrm{~m} \mathrm{~s}^{-1}\)
Relative separation between the particles \(s_{B, A}=v_{B, A} \times t\)
Hence, required time, \(t=\frac{s_{B A}}{v_{B d}}=\frac{150}{50}=3 \mathrm{~s}\)

A ball is thrown downwards with a speed of \(20 \mathrm{~ms}^{-1}\) from the top of a building \(150…

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