Q.45 A ball dropped from a bridge hits the surface of the water in 3 π . The height of
the bridge, ignoring air resistance, is __________ π (rounded off to one
decimal place).
(Use π = 9.8 ππ β2)
Understanding Free Fall Motion
Free fall occurs when an object moves solely under gravityβs influence, experiencing constant downward acceleration of 9.8 m/sΒ² (neglecting air resistance). When a ball dropped from a bridge hits water in exactly 3 seconds, kinematic equations reveal the bridge height precisely .
The core equation for free fall from rest is \( h = \frac{1}{2} g t^2 \), where initial velocity u = 0, making calculations straightforward for CSIR NET physics problems .
Problem Data Analysis
- Time: \( t = 3 \) s
- Gravity: \( g = 9.8 \) m/sΒ²
- Initial velocity: \( u = 0 \) m/s (dropped, not thrown)
- Find: Bridge height \( h \) (to 1 decimal place)
Key insight: βDroppedβ means zero initial velocity, eliminating the ut term from kinematic equations .
Step-by-Step Calculation
- \( t^2 = 9 \) sΒ²
- \( h = 4.9 \times 9 = 44.1 \) m
Why Other Answers Fail
29.4 m: Forgot Β½ factor (\( h = g t^2 \) error)
88.2 m: Doubled correct answer (used \( h = g t^2 \))
22.05 m: Divided by 2 twice
49.0 m: Used \( g = 10 \) m/sΒ² approximation
Only 44.1 m correctly applies \( \frac{1}{2} g t^2 \) with precise values .
Verification Methods
Check final velocity: \( v = g t = 9.8 \times 3 = 29.4 \) m/s
Using third equation: \( v^2 = 2 g h = 2 \times 9.8 \times 44.1 = 864.36 \), so \( v = \sqrt{864.36} = 29.4 \) m/s β
Reverse check: \( t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{88.2}{9.8}} = 3 \) s β .
Common Exam Mistakes
- Mistake 1: Ignoring Β½ coefficient β double answer
- Mistake 2: Using \( g = 10 \) instead of 9.8 m/sΒ²
- Mistake 3: Assuming thrown (u β 0)
- Mistake 4: Wrong precision (not 1 decimal place)
CSIR NET tip: Always match given g value and verify with multiple equations .
Real-World Applications
- Bridge safety analysis (fall hazards)
- Infrastructure design calculations
- Physics lab dropped-object experiments
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