Explanation:
B Given,
\(\mathrm{v}=0, \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{-2}, \mathrm{t}=1 \mathrm{sec}\).
From first equation of motion, we get-
\(\mathrm{v}=\mathrm{u}-\mathrm{gt}\)
\(0=\mathrm{u}-10 \times 1\)
\(\mathrm{u}=10 \mathrm{~m} / \mathrm{s}\)
By using third equation of motion, we get
\(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gH}\)
\(0=\mathrm{u}^{2}-2 \mathrm{gH}\)
Height \((\mathrm{H})\), to which the ball will rise is,
\(\mathrm{H}=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}\)
\(\mathrm{H}=\frac{(10)^{2}}{2 \times 10}\)
\(\mathrm{H}=5 \mathrm{~m}\)