Question

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

• A. 24.7 m
• B. 19.8 m
• C. 9.8 m
• D. 4.9 m

Answer 1

Answer: (C)

9.8 m

Answer 2

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
$s =u t+\frac{1}{2} g t^{2}$
$=-4.9 \times 2+\frac{1}{2} \times 9.8 \times(2)^{2}$
$=-9.8+19.6=9.8\, m$