Question

A boy throws a ball vertically upwards from a bridge with velocity 5 m/s. It strikes water surface after 2 s. The height of the bridge is (Take g = 10 m/s$^2$) ______.

• A. 20 m
• B. 15 m
• C. 12 m
• D. 10 m

Hint:

Always use displacement ($S$), not distance, in kinematic equations! If an object lands lower than it started, $S$ MUST be negative. This prevents having to break the trip into "up" and "down" chunks.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We must find the height of a bridge. A ball is thrown upwards, reverses direction, and falls to the water below. This is a 1D kinematics problem with constant acceleration.

Step 2: Detailed Explanation:
Let's establish a strict sign convention. Let the upward direction be positive ($+$) and the downward direction be negative ($-$).
- Initial velocity ($u$) = $+5$ m/s (thrown upwards).
- Acceleration due to gravity ($g$) = $-10$ m/s$^2$ (acts downwards).
- Total time of flight ($t$) = $2$ s.
- Displacement ($S$) = The final position relative to the starting point. Since the water is below the bridge, $S = -h$, where $h$ is the height of the bridge.
Apply the second equation of motion:
$S = ut + \frac{1}{2}at^2$
Substitute the known values:
$-h = (5)(2) + \frac{1}{2}(-10)(2)^2$
$-h = 10 - 5(4)$
$-h = 10 - 20$
$-h = -10 \text{ m}$
Multiply by $-1$:
$h = 10 \text{ m}$
The height of the bridge is 10 meters.

Step 3: Final Answer:
The height of the bridge is 10 m, matching option (d).