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  4. A ball falls from height h. After 1 second, another ball falls freely from a point 20 m below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h :- (g = 10 m/s)

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Q. A ball falls from height $h$. After $1$ second, another ball falls freely from a point $20\, m$ below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h :- $(g = 10\, m/s)$

Solution:

For first ball $h =\frac{1}{2} gt ^{2}$ ...(i)
For second ball $( h -20)=\frac{1}{2} g ( t -1)^{2}$ ...(ii)
image
subtract equation (ii) from (i)
$20=-5+10 t \Rightarrow \therefore t =2.5\, \sec$
Hence, height $h =\frac{1}{2} \times 10 \times(2.5)^{2}=31.2\, m$