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 Concept:
We use the second equation of motion for constant acceleration. We must be careful with the sign convention: upward is positive, downward is negative.

Step 2: Formula Application:
$s = ut + \frac{1}{2} at^2$.

Step 3: Explanation:
Let the bridge level be $y = 0$.
$u = +5$ m/s (upward), $a = -g = -10$ m/s², $t = 2$ s.
$s = (5)(2) + \frac{1}{2}(-10)(2)^2$
$s = 10 - 20 = -10$ m.
The negative sign indicates the displacement is 10 m below the starting point (the bridge).

Step 4: Final Answer:
The height of the bridge is 10 m.