A stone is thrown with an initial speed of $4.9 ext{ m/s}$ from a bridge in vertically upward direction. It

Question

A stone is thrown with an initial speed of $4.9 	ext{ m/s}$ from a bridge in vertically upward direction. It falls down in water after $2 	ext{ sec}$. The height of the bridge is:

Accepted Answer

1. **Sign Convention:** Let the upward direction be positive ($+$) and the downward direction be negative ($-$).
   - Initial velocity ($u$) = $+4.9 	ext{ m/s}$ (upwards).
   - Acceleration ($a$) = $-g = -9.8 	ext{ m/s}^2$ (acting downwards).
   - Time taken ($t$) = $2 	ext{ s}$.
   - Displacement ($s$) = $-h$ (where $h$ is the height of the bridge, since the final position is below the starting point).
2. **Apply Kinematic Equation:** Use the equation $s = ut + \frac{1}{2}at^2$ .
   $$ -h = 4.9(2) + \frac{1}{2}(-9.8)(2)^2 $$
   $$ -h = 9.8 + (-4.9)(4) $$
   $$ -h = 9.8 - 19.6 $$
   $$ -h = -9.8 $$
3. **Calculate Height:**
   $$ h = 9.8 	ext{ m} $$

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