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Q.
A stone is thrown with an initial speed of $4.9\, m / s$ from a bridge in vertically upward direction. It falls down in water after $2\, \sec$. The height of bridge is
Here: Initial velocity $u=-4.9\, m / s$ (-ve is due to vertically upward motion)
Total time $t=2\, \sec$
The height of the bridge is given by
$s =u t+\frac{1}{2} g t^{2}$
$=-4.9 \times 2+\frac{1}{2} \times 9.8 \times(2)^{2}$
$=-9.8+19.6=9.8\, m$